The document proves mathematical statements using induction. It shows: 1) Induction can be used to prove the summation formula 1 + 2 + 3 + ... + n = n(n+1)/2. It shows the base case of n=1 is true, and assumes the formula is true for n=k to prove it is true for n=k+1. 2) Induction can prove inequalities, like n < 2n for all positive integers n. It shows the base case is true and assumes the statement is true for n=k to prove it is true for n=k+1. 3) Induction can prove divisibility properties, like n3 - n being

1. 1 + 3 + 5 +....+ (2n - 1) = n²


2.  Now the LHS of P(k + 1) is
1 + 3 + 5 + ….. + (2k + 1)
= 1 + 3 + 5 +…. + (2k-1) + (2k + 1)
 by making the next-to-last term explicit
= k² + (2k + 1) by inductive hypothesis.

3.  And the RHS of P(k + 1) is(k + 1)² = k²+ 2k + 1 by basic
algebra.
k² + (2k + 1) = k²+ 2k + 1
 So the left-hand and right-hand sides of P(k+1) equal the
same quantity, and thus and thus P(k+1)
 is true [as was to be shown].

4. I
NDUCTION
Example : HYPOTHESIS
4 + 7 + 10 +- - - - + 3n+2= ½(3n² +5n ) prove by inductive.
-assume it is true ( hypothesis ) and then
If n= n+1 the sum should be ½ [3(n+1)² +5(n+1). prove it
-The term of n+1 = 3(n+1)+2
-The sum of (n+1) = ½(3n² +5n )+ 3(n+1)+2
½(3n² +5n + 6n+6+4)
= ½ [(3n² +6n + 3) + 5n+5
= ½[3(n+1)²+ 5(n+1)]
---> if inductive hypothesis P(n) is true , then P(n+1) must be true

5. MATHEMATICAL INDUCTION TO
CONSTRUCT PROOFS INVOLVING
FORMULAS, DIVISIBILITY PROPERTIES
AND INEQUALITIES

6. FORMULA : SUMMATION FORMULAE
 Problem 1:
Use mathematical induction to prove that
1 + 2 + 3 + ... + n = n (n + 1) / 2
for all positive integers n.
 Solution to Problem 1:
 Let the statement P (n) be
1 + 2 + 3 + ... + n = n (n + 1) / 2
STEP 1: We first show that p (1) is true.
Left Side = 1
Right Side = 1 (1 + 1) / 2 = 1
Both sides of the statement are equal hence p (1) is true.

7.  STEP 2: We now assume that p (k) is true
1 + 2 + 3 + ... + k = k (k + 1) / 2
 and show that p (k + 1) is true by adding k + 1 to both sides of
the above statement
1 + 2 + 3 + ... + k + (k + 1) = k (k + 1) / 2 + (k + 1)
= (k + 1)(k + 2) / 2
 The last statement may be written as
1 + 2 + 3 + ... + k + (k + 1) = (k + 1)(k + 2) / 2
 Which is the statement p(k + 1).

8.  PROVING INEQUALITIES
n<2ⁿ
For all positive integer n
Let p(n)=n<2ⁿ
Basis step:P(1) is true
1<2¹=2
Inductive step: p(k) is true, then p(k+1) is true
P(k)= k<2ᴷ
P(k+1)= k+1<2ᴷ ¹ ᴷ
Add 1 to both sides
K+1 < 2ᴷ+1 <= 2ᴷ+2ᴷ =2.2ᴷ =2ᴷ
ᴷ¹
P(k+1) is true

9.  PROVING DIVISIBILITY
Prove n³-n is divisible by 3. n is positve integer
Basis step: p(1) is true
1³-1=0 is divisible by 3
Inductive step: p(k) is true
P(k)= k³-k is divisible by 3
P(k+1)= (k+1) ³-(k+1) is divisible by 3
P(k+1) ³-(k+1 ) = (k³+3k²+3k+1)- (k+1 )
(k³-k)+3(k²+k)