SlideShare a Scribd company logo
1 of 53
Download to read offline
Mathematical Induction
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
Test: n = 1
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1
Test: n = 1       L.H .S  2
                          1
                         1
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                             1
Test: n = 1       L.H .S  2                R.H .S  2 
                          1                              1
                         1                        1
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                                  1
Test: n = 1       L.H .S  2                     R.H .S  2 
                          1                                   1
                         1                             1
                                L.H .S  R.H .S
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                                  1
Test: n = 1       L.H .S  2                     R.H .S  2 
                          1                                   1
                         1                             1
                                L.H .S  R.H .S

               1 1         1       1
A n  k  1  2  2    2  2 
              2 3         k        k
Mathematical Induction
                    1 1         1       1
e.g. i  Prove 1  2  2    2  2 
                   2 3         n        n
                           1                                  1
Test: n = 1       L.H .S  2                     R.H .S  2 
                          1                                   1
                         1                             1
                                L.H .S  R.H .S

               1 1         1       1
A n  k  1  2  2    2  2 
              2 3         k        k
                  1 1            1            1
P n  k  1 1  2  2              2
                 2 3          k  12
                                            k 1
Proof:
   1 1          1          1 1       1   1
1 2  2            1 2  2  2 
  2 3        k  12
                          2 3       k k  12
Proof:
   1 1          1          1 1        1    1
1 2  2            1 2  2  2 
  2 3        k  12
                          2 3         k k  12
                           1    1
                       2 
                           k k  12
Proof:
   1 1          1          1 1         1    1
1 2  2            1 2  2  2 
  2 3        k  12
                           2 3         k k  12
                           1       1
                       2 
                           k k  12
                           k  1  k
                                  2
                       2
                            k k  1
                                     2
Proof:
   1 1          1          1 1          1    1
1 2  2            1 2  2  2 
  2 3        k  12
                           2 3          k k  12
                           1        1
                       2 
                           k k  12
                           k  1  k
                                   2
                       2
                            k k  1
                                      2

                            k 2  k 1
                       2
                            k k  1
                                      2
Proof:
   1 1          1          1 1            1      1
1 2  2            1 2  2  2 
  2 3        k  12
                           2 3            k k  12
                           1        1
                       2 
                           k k  12
                           k  1  k
                                   2
                       2
                            k k  1
                                      2

                            k 2  k 1
                       2
                            k k  1
                                      2


                             k2  k        1
                       2              
                            k k  1 k k  1
                                     2         2
Proof:
   1 1          1          1 1            1      1
1 2  2            1 2  2  2 
  2 3        k  12
                           2 3            k k  12
                           1        1
                       2 
                           k k  12
                           k  1  k
                                   2
                       2
                             k k  1
                                      2

                            k 2  k 1
                       2
                            k k  1
                                      2


                              k2  k       1
                       2              
                            k k  1 k k  1
                                     2         2

                            k k  1
                       2
                            k k  1
                                     2
Proof:
   1 1          1           1 1            1      1
1 2  2            1 2  2  2 
  2 3        k  12
                           2 3             k k  12
                            1        1
                       2 
                            k k  12
                            k  1  k
                                    2
                       2
                              k k  1
                                       2

                             k 2  k 1
                       2
                             k k  1
                                       2


                               k2  k       1
                       2               
                             k k  1 k k  1
                                      2         2

                             k k  1
                       2
                             k k  1
                                      2

                                  1
                          2
                                k 1
Proof:
   1 1           1           1 1            1      1
1 2  2             1 2  2  2 
  2 3         k  12
                            2 3             k k  12
                             1        1
                        2 
                             k k  12
                             k  1  k
                                     2
                        2
                               k k  1
                                        2

                              k 2  k 1
                         2
                              k k  1
                                        2


                                k2  k       1
                        2               
                              k k  1 k k  1
                                       2         2

                              k k  1
                        2
                              k k  1
                                       2

                                   1
                           2
                                 k 1
     1 1            1              1
1  2  2              2
     2 3         k  12
                                 k 1
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
Test: n = 1
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
Test: n = 1   a1  2  2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
   Show that an  2 for n  1
Test: n = 1   a1  2  2
A n  k  ak  2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  ak  2

P   n  k  1 ak 1  2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  ak  2

P   n  k  1 ak 1  2
Proof:
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  ak  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  ak  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
           22
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  ak  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
           22
           4
          2
(ii) A sequence is defined by;
                a1  2         an1  2  an for n  1
    Show that an  2 for n  1
Test: n = 1    a1  2  2
A n  k  ak  2

P   n  k  1 ak 1  2
Proof:
      ak 1  2  ak
           22
             4
            2
     ak 1  2
iii The sequences xn and yn are defined by;
                                           xn  yn         2x y
           x1  5, y1  2         xn1            , yn1  n n
                                              2            xn  yn
     Prove xn yn  10 for n  1
iii The sequences xn and yn are defined by;
                                           xn  yn         2x y
              x1  5, y1  2      xn1            , yn1  n n
                                              2            xn  yn
     Prove xn yn  10 for n  1
Test: n = 1
iii The sequences xn and yn are defined by;
                                            xn  yn         2x y
            x1  5, y1  2         xn1            , yn1  n n
                                               2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                     10
iii The sequences xn and yn are defined by;
                                            xn  yn         2x y
            x1  5, y1  2         xn1            , yn1  n n
                                               2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                    10
A n  k  xk yk  10
iii The sequences xn and yn are defined by;
                                            xn  yn         2x y
             x1  5, y1  2        xn1            , yn1  n n
                                               2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                    10
A n  k  xk yk  10

P   n  k  1 xk 1 yk 1  10
iii The sequences xn and yn are defined by;
                                            xn  yn         2x y
             x1  5, y1  2        xn1            , yn1  n n
                                               2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                    10
A n  k  xk yk  10

P   n  k  1 xk 1 yk 1  10
Proof:
iii The sequences xn and yn are defined by;
                                                     xn  yn         2x y
               x1  5, y1  2               xn1            , yn1  n n
                                                        2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                    10
A n  k  xk yk  10

P     n  k  1 xk 1 yk 1  10
Proof:
                   xk  yk  2 xk yk 
    xk 1 yk 1           
                             x  y    
                   2  k            k 
iii The sequences xn and yn are defined by;
                                                     xn  yn         2x y
               x1  5, y1  2               xn1            , yn1  n n
                                                        2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                    10
A n  k  xk yk  10

P     n  k  1 xk 1 yk 1  10
Proof:
                   xk  yk  2 xk yk 
    xk 1 yk 1           
                             x  y    
                   2  k            k 

               xk y k
               10
iii The sequences xn and yn are defined by;
                                                     xn  yn         2x y
               x1  5, y1  2               xn1            , yn1  n n
                                                        2            xn  yn
      Prove xn yn  10 for n  1
Test: n = 1 x1 y1  52
                    10
A n  k  xk yk  10

P     n  k  1 xk 1 yk 1  10
Proof:
                   xk  yk  2 xk yk 
    xk 1 yk 1           
                             x  y    
                   2  k            k 

               xk y k
               10
 xk 1 yk 1  10
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2
  L.H .S  a1
        1
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                          1  5 
  L.H .S  a1                    R.H .S        
                                           2 
        1
                                         1.62
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S
  L.H .S  a2
         1
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S              2
                                       1  5 
  L.H .S  a2                 R.H .S        
                                        2 
         1
                                      2.62
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                             n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                             1
                                       1  5 
  L.H .S  a1                 R.H .S        
                                        2 
        1
                                      1.62
                  L.H .S  R.H .S              2
                                       1  5 
  L.H .S  a2                 R.H .S        
                                        2 
         1
                                      2.62
                  L.H .S  R.H .S
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                                n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                                 1
                                            1  5 
  L.H .S  a1                     R.H .S         
                                             2 
         1
                                          1.62
                  L.H .S  R.H .S                    2
                                             1  5 
  L.H .S  a2                      R.H .S         
                                              2 
         1
                                           2.62
                   L.H .S  R.H .S
                                          k 1                  k
                                 1  5               1  5 
  A n  k  1 & n  k  ak 1          & ak             
                                  2                   2 
(iv) The Fibonacci sequence is defined by;
              a1  a2  1       an1  an  an1 for n  1
                                n
                     1  5 
     Prove that an         for n  1
                      2 
Test: n = 1 and n =2                                 1
                                            1  5 
  L.H .S  a1                     R.H .S         
                                             2 
         1
                                          1.62
                  L.H .S  R.H .S                    2
                                             1  5 
  L.H .S  a2                      R.H .S         
                                              2 
         1
                                           2.62
                   L.H .S  R.H .S
                                          k 1                  k
                                 1  5               1  5 
  A n  k  1 & n  k  ak 1          & ak             
                                  2                   2 
                                    k 1
                       1  5 
 P n  k  1 ak 1        
                        2 
Proof:   ak 1  ak  ak 1
Proof:   ak 1  ak  ak 1
                              k    k 1
               1  5  1  5 
                          
                2   2 
Proof:   ak 1  ak  ak 1
                              k     k 1
               1  5  1       5
                              
                2   2           
                        k 1         1         2
                1  5   1    5    1  5  
                                        
                2   2              2      
Proof:   ak 1  ak  ak 1
                              k   k 1
               1  5  1 5
                        
                2   2     
                        k 1   1         2
                1  5   1 
                            5    1  5  
                                  
                2   2    
                                  2      
                       k 1
               1  5   2       4 
                                 2
                2  1  5 1  5  
Proof:   ak 1  ak  ak 1
                              k                k 1
               1  5  1        5
                               
                2   2            
                        k 1          1         2
                1  5   1     5    1  5  
                                         
                2   2               2      
                           k 1
               1      5  2           4 
                                        2
                2         1  5 1  5  
                           k 1
               1      5  2  2 5  4
                                  2 
                2          1  5  
                              k 1
               1  5               62 5 
                                           2
                2                   1  5  
Proof:   ak 1  ak  ak 1
                              k       k 1
               1  5  1        5
                               
                2   2            
                        k 1          1         2
                1  5   1     5    1  5  
                                         
                2   2               2      
                           k 1
               1      5  2           4 
                                        2
                2         1  5 1  5  
                           k 1
               1      5  2  2 5  4
                                  2 
                2          1  5  
                              k 1
               1  5   6  2 5 
                               
                  2   1  5 2 
                        k 1
               1  5 
                    
                2 
Proof:     ak 1  ak  ak 1
                                k       k 1
                 1  5  1        5
                                 
                  2   2            
                          k 1          1         2
                  1  5   1     5    1  5  
                                           
                  2   2               2      
                             k 1
                 1      5  2           4 
                                          2
                  2         1  5 1  5  
                             k 1
                 1      5  2  2 5  4
                                    2 
                  2          1  5  
                                k 1
                   1  5   6  2 5 
                                   
                      2   1  5 2 
                            k 1
                   1  5 
                        
                    2 
                            k 1
                   1  5 
          ak 1        
                    2 
Proof:           ak 1  ak  ak 1
                                      k       k 1
                       1  5  1        5
                                       
                        2   2            
                                k 1          1         2
                        1  5   1     5    1  5  
                                                 
                        2   2               2      
                                   k 1
                       1      5  2           4 
     Sheets                                     2
                        2         1  5 1  5  
                                   k 1
         +             1      5  2  2 5  4
                                          2 
                        2          1  5  
 Exercise 10E*
                                      k 1
                        1  5   6  2 5 
                                        
                           2   1  5 2 
                                 k 1
                        1  5 
                             
                         2 
                                 k 1
                        1  5 
               ak 1        
                         2 

More Related Content

Viewers also liked

Women, Baldness, and Hair Loss Treatments
Women, Baldness, and Hair Loss TreatmentsWomen, Baldness, and Hair Loss Treatments
Women, Baldness, and Hair Loss Treatmentsstewart_bay
 
How to Hit a Home Run in Your Venue
How to Hit a Home Run in Your VenueHow to Hit a Home Run in Your Venue
How to Hit a Home Run in Your VenueStanley Sarpong
 
Izbor učenika generacije šk. 2011./2012. godine
Izbor učenika generacije šk. 2011./2012. godineIzbor učenika generacije šk. 2011./2012. godine
Izbor učenika generacije šk. 2011./2012. godineosem_present
 
Vulcan entre légende et modernité - présentation
Vulcan  entre légende et modernité - présentationVulcan  entre légende et modernité - présentation
Vulcan entre légende et modernité - présentationMarinela
 
Axxiome / 2innovate IT - Mobile Offering - 20120706
Axxiome / 2innovate IT - Mobile Offering - 20120706Axxiome / 2innovate IT - Mobile Offering - 20120706
Axxiome / 2innovate IT - Mobile Offering - 20120706Ivan Kechichian Maggio
 
Proyecto pre
Proyecto pre Proyecto pre
Proyecto pre rociosito
 
Domingo de ramos fotos
Domingo de ramos fotosDomingo de ramos fotos
Domingo de ramos fotosAna Marisa
 
Domingo 6º del to. ciclo b. día 12 de febrero del 2012.
Domingo 6º del to. ciclo b. día 12 de febrero del 2012. Domingo 6º del to. ciclo b. día 12 de febrero del 2012.
Domingo 6º del to. ciclo b. día 12 de febrero del 2012. FEDERICO ALMENARA CHECA
 
Syllabus lab 1er semestre
Syllabus lab 1er semestreSyllabus lab 1er semestre
Syllabus lab 1er semestreusssec1
 
Guía cap a una mallorca per a tothom
Guía cap a una mallorca per a tothomGuía cap a una mallorca per a tothom
Guía cap a una mallorca per a tothomPlatges Accessibles
 
Quirky productswithintro
Quirky productswithintroQuirky productswithintro
Quirky productswithintrojoergenpleijte
 
How to install_kitchen_cabinet
How to install_kitchen_cabinetHow to install_kitchen_cabinet
How to install_kitchen_cabinetBaha Architecture
 
XS Games Postcard [June 1999]
XS Games Postcard [June 1999]XS Games Postcard [June 1999]
XS Games Postcard [June 1999]Kimberly L. King
 

Viewers also liked (16)

Women, Baldness, and Hair Loss Treatments
Women, Baldness, and Hair Loss TreatmentsWomen, Baldness, and Hair Loss Treatments
Women, Baldness, and Hair Loss Treatments
 
How to Hit a Home Run in Your Venue
How to Hit a Home Run in Your VenueHow to Hit a Home Run in Your Venue
How to Hit a Home Run in Your Venue
 
Xcvxcvwxcv
XcvxcvwxcvXcvxcvwxcv
Xcvxcvwxcv
 
Izbor učenika generacije šk. 2011./2012. godine
Izbor učenika generacije šk. 2011./2012. godineIzbor učenika generacije šk. 2011./2012. godine
Izbor učenika generacije šk. 2011./2012. godine
 
Vulcan entre légende et modernité - présentation
Vulcan  entre légende et modernité - présentationVulcan  entre légende et modernité - présentation
Vulcan entre légende et modernité - présentation
 
Axxiome / 2innovate IT - Mobile Offering - 20120706
Axxiome / 2innovate IT - Mobile Offering - 20120706Axxiome / 2innovate IT - Mobile Offering - 20120706
Axxiome / 2innovate IT - Mobile Offering - 20120706
 
Proyecto pre
Proyecto pre Proyecto pre
Proyecto pre
 
Domingo de ramos fotos
Domingo de ramos fotosDomingo de ramos fotos
Domingo de ramos fotos
 
Domingo 6º del to. ciclo b. día 12 de febrero del 2012.
Domingo 6º del to. ciclo b. día 12 de febrero del 2012. Domingo 6º del to. ciclo b. día 12 de febrero del 2012.
Domingo 6º del to. ciclo b. día 12 de febrero del 2012.
 
Syllabus lab 1er semestre
Syllabus lab 1er semestreSyllabus lab 1er semestre
Syllabus lab 1er semestre
 
Guía cap a una mallorca per a tothom
Guía cap a una mallorca per a tothomGuía cap a una mallorca per a tothom
Guía cap a una mallorca per a tothom
 
Quirky productswithintro
Quirky productswithintroQuirky productswithintro
Quirky productswithintro
 
Dona carolina site
Dona carolina siteDona carolina site
Dona carolina site
 
How to install_kitchen_cabinet
How to install_kitchen_cabinetHow to install_kitchen_cabinet
How to install_kitchen_cabinet
 
Proyecto Regiones Devastadas
Proyecto Regiones DevastadasProyecto Regiones Devastadas
Proyecto Regiones Devastadas
 
XS Games Postcard [June 1999]
XS Games Postcard [June 1999]XS Games Postcard [June 1999]
XS Games Postcard [June 1999]
 

Similar to X2 t08 02 induction (2012)

11X1 T10 10 mathematical induction 3
11X1 T10 10 mathematical induction 311X1 T10 10 mathematical induction 3
11X1 T10 10 mathematical induction 3Nigel Simmons
 
11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)Nigel Simmons
 
11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)Nigel Simmons
 
11X1 T14 10 mathematical induction 3 (2010)
11X1 T14 10 mathematical induction 3 (2010)11X1 T14 10 mathematical induction 3 (2010)
11X1 T14 10 mathematical induction 3 (2010)Nigel Simmons
 
11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)Nigel Simmons
 
11X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 211X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 2Nigel Simmons
 
11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)Nigel Simmons
 
X2 t08 02 induction (2013)
X2 t08 02 induction (2013)X2 t08 02 induction (2013)
X2 t08 02 induction (2013)Nigel Simmons
 
mathematical induction and stuff Induction.pptx
mathematical induction and stuff Induction.pptxmathematical induction and stuff Induction.pptx
mathematical induction and stuff Induction.pptxZenLooper
 
mathematical induction
mathematical inductionmathematical induction
mathematical inductionankush_kumar
 
11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)Nigel Simmons
 
mathematical induction
mathematical inductionmathematical induction
mathematical inductionankush_kumar
 
11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)Nigel Simmons
 
11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)Nigel Simmons
 

Similar to X2 t08 02 induction (2012) (15)

11X1 T10 10 mathematical induction 3
11X1 T10 10 mathematical induction 311X1 T10 10 mathematical induction 3
11X1 T10 10 mathematical induction 3
 
11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)11 x1 t14 10 mathematical induction 3 (2012)
11 x1 t14 10 mathematical induction 3 (2012)
 
11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)11X1 T14 10 mathematical induction 3 (2011)
11X1 T14 10 mathematical induction 3 (2011)
 
11X1 T14 10 mathematical induction 3 (2010)
11X1 T14 10 mathematical induction 3 (2010)11X1 T14 10 mathematical induction 3 (2010)
11X1 T14 10 mathematical induction 3 (2010)
 
11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)11 x1 t14 10 mathematical induction 3 (2013)
11 x1 t14 10 mathematical induction 3 (2013)
 
11X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 211X1 T10 09 mathematical induction 2
11X1 T10 09 mathematical induction 2
 
2.4 edited1
2.4 edited12.4 edited1
2.4 edited1
 
11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)11 x1 t14 08 mathematical induction 1 (2013)
11 x1 t14 08 mathematical induction 1 (2013)
 
X2 t08 02 induction (2013)
X2 t08 02 induction (2013)X2 t08 02 induction (2013)
X2 t08 02 induction (2013)
 
mathematical induction and stuff Induction.pptx
mathematical induction and stuff Induction.pptxmathematical induction and stuff Induction.pptx
mathematical induction and stuff Induction.pptx
 
mathematical induction
mathematical inductionmathematical induction
mathematical induction
 
11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)11X1 T14 09 mathematical induction 2 (2011)
11X1 T14 09 mathematical induction 2 (2011)
 
mathematical induction
mathematical inductionmathematical induction
mathematical induction
 
11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)11 x1 t14 09 mathematical induction 2 (2012)
11 x1 t14 09 mathematical induction 2 (2012)
 
11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)11X1 T14 09 mathematical induction 2 (2010)
11X1 T14 09 mathematical induction 2 (2010)
 

More from Nigel Simmons

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATENigel Simmons
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)Nigel Simmons
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)Nigel Simmons
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)Nigel Simmons
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)Nigel Simmons
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)Nigel Simmons
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)Nigel Simmons
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)Nigel Simmons
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)Nigel Simmons
 

More from Nigel Simmons (20)

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
 

Recently uploaded

Concept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.CompdfConcept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.CompdfUmakantAnnand
 
The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13Steve Thomason
 
mini mental status format.docx
mini    mental       status     format.docxmini    mental       status     format.docx
mini mental status format.docxPoojaSen20
 
Crayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon ACrayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon AUnboundStockton
 
ECONOMIC CONTEXT - LONG FORM TV DRAMA - PPT
ECONOMIC CONTEXT - LONG FORM TV DRAMA - PPTECONOMIC CONTEXT - LONG FORM TV DRAMA - PPT
ECONOMIC CONTEXT - LONG FORM TV DRAMA - PPTiammrhaywood
 
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111Sapana Sha
 
Mastering the Unannounced Regulatory Inspection
Mastering the Unannounced Regulatory InspectionMastering the Unannounced Regulatory Inspection
Mastering the Unannounced Regulatory InspectionSafetyChain Software
 
Paris 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activityParis 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activityGeoBlogs
 
The basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxThe basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxheathfieldcps1
 
call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️
call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️
call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️9953056974 Low Rate Call Girls In Saket, Delhi NCR
 
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxContemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxRoyAbrique
 
Science 7 - LAND and SEA BREEZE and its Characteristics
Science 7 - LAND and SEA BREEZE and its CharacteristicsScience 7 - LAND and SEA BREEZE and its Characteristics
Science 7 - LAND and SEA BREEZE and its CharacteristicsKarinaGenton
 
Introduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxIntroduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxpboyjonauth
 
Sanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfSanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfsanyamsingh5019
 
Class 11 Legal Studies Ch-1 Concept of State .pdf
Class 11 Legal Studies Ch-1 Concept of State .pdfClass 11 Legal Studies Ch-1 Concept of State .pdf
Class 11 Legal Studies Ch-1 Concept of State .pdfakmcokerachita
 
POINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptx
POINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptxPOINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptx
POINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptxSayali Powar
 
Employee wellbeing at the workplace.pptx
Employee wellbeing at the workplace.pptxEmployee wellbeing at the workplace.pptx
Employee wellbeing at the workplace.pptxNirmalaLoungPoorunde1
 

Recently uploaded (20)

Concept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.CompdfConcept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.Compdf
 
The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13
 
Model Call Girl in Bikash Puri Delhi reach out to us at 🔝9953056974🔝
Model Call Girl in Bikash Puri  Delhi reach out to us at 🔝9953056974🔝Model Call Girl in Bikash Puri  Delhi reach out to us at 🔝9953056974🔝
Model Call Girl in Bikash Puri Delhi reach out to us at 🔝9953056974🔝
 
9953330565 Low Rate Call Girls In Rohini Delhi NCR
9953330565 Low Rate Call Girls In Rohini  Delhi NCR9953330565 Low Rate Call Girls In Rohini  Delhi NCR
9953330565 Low Rate Call Girls In Rohini Delhi NCR
 
mini mental status format.docx
mini    mental       status     format.docxmini    mental       status     format.docx
mini mental status format.docx
 
Crayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon ACrayon Activity Handout For the Crayon A
Crayon Activity Handout For the Crayon A
 
ECONOMIC CONTEXT - LONG FORM TV DRAMA - PPT
ECONOMIC CONTEXT - LONG FORM TV DRAMA - PPTECONOMIC CONTEXT - LONG FORM TV DRAMA - PPT
ECONOMIC CONTEXT - LONG FORM TV DRAMA - PPT
 
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111Call Girls in Dwarka Mor Delhi Contact Us 9654467111
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
 
Mastering the Unannounced Regulatory Inspection
Mastering the Unannounced Regulatory InspectionMastering the Unannounced Regulatory Inspection
Mastering the Unannounced Regulatory Inspection
 
Staff of Color (SOC) Retention Efforts DDSD
Staff of Color (SOC) Retention Efforts DDSDStaff of Color (SOC) Retention Efforts DDSD
Staff of Color (SOC) Retention Efforts DDSD
 
Paris 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activityParis 2024 Olympic Geographies - an activity
Paris 2024 Olympic Geographies - an activity
 
The basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxThe basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptx
 
call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️
call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️
call girls in Kamla Market (DELHI) 🔝 >༒9953330565🔝 genuine Escort Service 🔝✔️✔️
 
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxContemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
 
Science 7 - LAND and SEA BREEZE and its Characteristics
Science 7 - LAND and SEA BREEZE and its CharacteristicsScience 7 - LAND and SEA BREEZE and its Characteristics
Science 7 - LAND and SEA BREEZE and its Characteristics
 
Introduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxIntroduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptx
 
Sanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfSanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdf
 
Class 11 Legal Studies Ch-1 Concept of State .pdf
Class 11 Legal Studies Ch-1 Concept of State .pdfClass 11 Legal Studies Ch-1 Concept of State .pdf
Class 11 Legal Studies Ch-1 Concept of State .pdf
 
POINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptx
POINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptxPOINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptx
POINT- BIOCHEMISTRY SEM 2 ENZYMES UNIT 5.pptx
 
Employee wellbeing at the workplace.pptx
Employee wellbeing at the workplace.pptxEmployee wellbeing at the workplace.pptx
Employee wellbeing at the workplace.pptx
 

X2 t08 02 induction (2012)

  • 2. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n
  • 3. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n Test: n = 1
  • 4. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 Test: n = 1 L.H .S  2 1 1
  • 5. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1
  • 6. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1  L.H .S  R.H .S
  • 7. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1  L.H .S  R.H .S 1 1 1 1 A n  k  1  2  2    2  2  2 3 k k
  • 8. Mathematical Induction 1 1 1 1 e.g. i  Prove 1  2  2    2  2  2 3 n n 1 1 Test: n = 1 L.H .S  2 R.H .S  2  1 1 1 1  L.H .S  R.H .S 1 1 1 1 A n  k  1  2  2    2  2  2 3 k k 1 1 1 1 P n  k  1 1  2  2     2 2 3 k  12 k 1
  • 9. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12
  • 10. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12
  • 11. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2
  • 12. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2
  • 13. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2
  • 14. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2 k k  1  2 k k  1 2
  • 15. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2 k k  1  2 k k  1 2 1  2 k 1
  • 16. Proof: 1 1 1 1 1 1 1 1 2  2   1 2  2  2  2 3 k  12 2 3 k k  12 1 1  2  k k  12 k  1  k 2  2 k k  1 2 k 2  k 1  2 k k  1 2 k2  k 1  2  k k  1 k k  1 2 2 k k  1  2 k k  1 2 1  2 k 1 1 1 1 1 1  2  2     2 2 3 k  12 k 1
  • 17. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1
  • 18. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1
  • 19. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2
  • 20. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2
  • 21. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2 P n  k  1 ak 1  2
  • 22. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2 P n  k  1 ak 1  2 Proof:
  • 23. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak
  • 24. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak  22
  • 25. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak  22  4 2
  • 26. (ii) A sequence is defined by; a1  2 an1  2  an for n  1 Show that an  2 for n  1 Test: n = 1 a1  2  2 A n  k  ak  2 P n  k  1 ak 1  2 Proof: ak 1  2  ak  22  4 2  ak 1  2
  • 27. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1
  • 28. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1
  • 29. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10
  • 30. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10 A n  k  xk yk  10
  • 31. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10
  • 32. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:
  • 33. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:  xk  yk  2 xk yk  xk 1 yk 1    x  y    2  k k 
  • 34. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:  xk  yk  2 xk yk  xk 1 yk 1    x  y    2  k k   xk y k  10
  • 35. iii The sequences xn and yn are defined by; xn  yn 2x y x1  5, y1  2 xn1  , yn1  n n 2 xn  yn Prove xn yn  10 for n  1 Test: n = 1 x1 y1  52  10 A n  k  xk yk  10 P n  k  1 xk 1 yk 1  10 Proof:  xk  yk  2 xk yk  xk 1 yk 1    x  y    2  k k   xk y k  10  xk 1 yk 1  10
  • 36. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2 
  • 37. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2
  • 38. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 L.H .S  a1 1
  • 39. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62
  • 40. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S
  • 41. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S L.H .S  a2 1
  • 42. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62
  • 43. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62  L.H .S  R.H .S
  • 44. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62  L.H .S  R.H .S k 1 k 1  5  1  5  A n  k  1 & n  k  ak 1    & ak     2   2 
  • 45. (iv) The Fibonacci sequence is defined by; a1  a2  1 an1  an  an1 for n  1 n 1  5  Prove that an    for n  1  2  Test: n = 1 and n =2 1 1  5  L.H .S  a1 R.H .S     2  1  1.62  L.H .S  R.H .S 2 1  5  L.H .S  a2 R.H .S     2  1  2.62  L.H .S  R.H .S k 1 k 1  5  1  5  A n  k  1 & n  k  ak 1    & ak     2   2  k 1 1  5  P n  k  1 ak 1     2 
  • 46. Proof: ak 1  ak  ak 1
  • 47. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5       2   2 
  • 48. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2   
  • 49. Proof: ak 1  ak  ak 1 k k 1 1  5  1 5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5   2 4      2  2  1  5 1  5  
  • 50. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4      2  2  1  5 1  5   k 1 1  5  2  2 5  4    2   2   1  5   k 1 1  5  62 5     2  2   1  5  
  • 51. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4      2  2  1  5 1  5   k 1 1  5  2  2 5  4    2   2   1  5   k 1 1  5   6  2 5       2   1  5 2  k 1 1  5     2 
  • 52. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4      2  2  1  5 1  5   k 1 1  5  2  2 5  4    2   2   1  5   k 1 1  5   6  2 5       2   1  5 2  k 1 1  5     2  k 1 1  5   ak 1     2 
  • 53. Proof: ak 1  ak  ak 1 k k 1 1  5  1  5      2   2  k 1 1 2  1  5   1  5 1  5           2   2    2    k 1 1  5  2 4  Sheets     2  2  1  5 1  5   k 1 + 1  5  2  2 5  4    2   2   1  5   Exercise 10E* k 1 1  5   6  2 5       2   1  5 2  k 1 1  5     2  k 1 1  5   ak 1     2 