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THE BINOMIAL THEOREM

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THE BINOMIAL THEOREM shows how to calculate a power of a binomial –
(x+ y)n -- without actually multiplying out.

For example, if we actually multiplied out the 4th power of (x + y) --
(x + y)4 = (x + y) (x + y) (x + y) (x + y)
-- then on collecting like terms we would find:
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 .   .  .  .  .  (1)

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THE BINOMIAL THEOREM

1. 1. Presented ByPresented ByGroup-JGroup-JMD. Habibur Rahman (41222009)MD. Habibur Rahman (41222009)Md. Obaidur Rahman Sikder (41222041)Md. Obaidur Rahman Sikder (41222041)Binomial TheoremBinomial Theorem
2. 2. A binomial is an algebraic expression containing 2 terms. For example, (x + y) is abinomial.We sometimes need to expand binomials as follows:(a + b)0= 1(a + b)1= a + b(a + b)2= a2+ 2ab + b2(a + b)3= a3+ 3a2b + 3ab2+ b3(a + b)4= a4+ 4a3b + 6a2b2+ 4ab3+ b4(a + b)5= a5+ 5a4b + 10a3b2+ 10a2b3+ 5ab4+ b5Clearly, doing this by direct multiplication gets quite tedious and can be ratherdifficult for larger powers or more complicated expressions.BinomialsBinomials
3. 3. In mathematics, Pascals triangle is a triangular array of the binomial coefficients in atriangle. It is named after the French mathematician Blaise Pascal in much of theWestern world, although other mathematicians studied it centuries before him in India,Greece, Iran, China, Germany, and Italy.Pascal’s trianglePascal’s triangle345610111111123410 51 11 1( )0a b 1+ =( )1a b 1a 1b+ = +( )2 2 2a b 1a 2ab 1b+ = + +( )3 3 2 2 3a b 1a 3a b 3ab 1b+ = + + +( )4 4 3 2 2 3 4a b 1a 4a b 6a b 4ab 1b+ = + + + +( )5 5 4 3 2 2 3 4 5a b 1a 5a b 10a b 10a b 5ab 1b+ = + + + + +
4. 4. 0C01C0 C11C22C33C44C55C21C31C41C51C52C53C54C43C32C422C03C04C05C0n n n 1r–1 r rc c c++ =Each row gives the combinatorial numbers, which are the binomial coefficients. Thatis, the row 1 2 1 are the combinatorial numbers 2Ck, which are the coefficients of(a + b)2. The next row, 1 3 3 1, are the coefficients of (a + b)3; and so on.
5. 5. Properties of binomial expansion (a+b)Properties of binomial expansion (a+b)nn( )n n n 0 n n–1 1 n n–2 2 n 1 n–1 n 0 n0 1 2 n–1 na b c a b c a b c a b ... c a b c a b+ = + + + + + Based on the binomial properties, the binomial theorem states that the followingbinomial formula is valid for all positive integer values of n. (a+b)nhas n+1 terms as 0 ≤ k ≤ n. The first term is anand the final term is bn Progressing from the first term to the last, the exponent of a decreases by fromterm to term (start at n and go down) while the exponent of b increases by (startat 0 and go up). The sum of the exponents of a and b in each term is n. If the coefficient of each term is multiplied by the exponent of a in that term, andthe product is divided by the number of that term, we obtain the coefficient of thenext term. Coefficients of terms equidistant from beginning and end is same as nck = ncn-k
6. 6. Example: What is (x+5)4Start withexponents:x450x351x252x153x054IncludeCoefficients:1x4504x3516x2524x1531x054Then write down the answer (including all calculations, such as 4×5, 6×52, etc):(x+5)4= x4+ 20x3+ 150x2+ 500x + 625 The nth row of the Pascals Triangle will be the coefficients of the expanded binomial. For each line, the number of products (i.e. the sum of the coefficients) is equal to 2n. For each line, the number of product groups is equal to n+1. . The binomial theorem can be applied to the powers of any binomial.For example, For a binomial involving subtraction, the theorem can be applied as long as the oppositeof the second term is used. This has the effect of changing the sign of every other termin the expansion:
7. 7. According to the theorem, it is possible to expand any power of x+ y into a sum of theformwhere each is a specific positive integer known as binomial coefficient. This formulais also referred to as the Binomial Formula or the Binomial Identity. Using summationnotation, it can be written asThe final expression follows from the previous one by the symmetry of x and y in the firstexpression, and by comparison it follows that the sequence of binomial coefficients in theformula is symmetrical.A variant of the binomial formula is obtained by substituting 1 for y, so that it involvesonly a single variable. In this form, the formula readsor equivalentlyStatement of the theoremStatement of the theorem
8. 8. THE BINOMIAL THEOREM shows how to calculate a power of a binomial –(x+ y)n-- without actually multiplying out.For example, if we actually multiplied out the 4th power of (x + y) --(x + y)4= (x + y) (x + y) (x + y) (x + y)-- then on collecting like terms we would find:(x + y)4= x4+ 4x3y + 6x2y2+ 4xy3+ y4. . . . . (1)Notice: The literal factors are all the combinations of a and b where the sum of theexponents is 4: x4, x3y, x2y2, xy3, y4.The degree of each term is 4.The first term is actually x4y0, which is x4· 1.Statement of the theoremStatement of the theorem
9. 9. Thus to "expand" (x + y)5, we would anticipate the following terms, in which the sum ofall the exponents is 5:(x + y)5= ? x5+ ? x4y + ? x3y2+ ? x2y3+ ? xy4+ ? y5The question is, What are the coefficients?They are called the binomial coefficients.(x + y)4= x4+ 4x3y + 6x2y2+ 4xy3+ y4In the expansion of (x + y)4, the binomial coefficients are 1 4 6 4 1 .Note the symmetry: The coefficients from left to right are the same right to left.The answer to the question, "What are the binomial coefficients?" is called the binomialtheorem. It shows how to calculate the coefficients in the expansion of (x+ y)n.Statement of the theoremStatement of the theorem
10. 10. The symbol for a binomial coefficient is . The upper index n is the exponent of theexpansion; the lower index k indicates which term, starting with k = 0.The lower index k is the exponent of y.For example, when n = 5, each term in the expansion of (x + y)5will look like this:x5 − kykk will successively take on the values 0 through 5.(x + y)5= x5+ x4y + x3y2+ x2y3+ xy4+ y5Again, each lower index is the exponent of y. The first term has k = 0, because in thefirst term, y appears as y0, which is 1.Statement of the theoremStatement of the theorem
11. 11. Now, what are these binomial coefficients, ?The theorem states that the binomial coefficients are none other than the combinatorialnumbers, nCk.= nCk(a + b)5 = 5C0a5+ 5C1a4b + 5C2a3b2+ 5C3a2b3+ 5C4ab4+ 5C5b5=1a5+ a4b + a3b2+ a2b3+ ab4+ b5=a5+ 5a4b + 10a3b2+ 10a2b3+ 5ab4+ b5The binomial coefficients here are 1 5 10 10 5 1.Statement of the theoremStatement of the theoremNote: The coefficient of the first term is 1, and the coefficient of the second term isthe same as the exponent of (a + b), which here is 5.
12. 12. Using sigma notation, and factorials for the combinatorial numbers, here is thebinomial theorem for (a+b)n:What follows the summation sign is the general term. Each term in the sum will looklike that -- the first term having k = 0; then k = 1, k = 2, and so on, up to k = n.Notice that the sum of the exponents (n − k) + k, always equals n.Statement of the theoremStatement of the theoremExample 1.a) The term a8b4occurs in the expansion of what binomial?Answer. (a + b)12. The sum of 8 + 4 is 12.b) In that expansion, what number is the coefficient of a8b4?Answer. It is the combinatorial number,12C4=12· 11· 10· 91· 2· 3· 4= 495Note: The lower index, in this case 4, is the exponent of b.This same number is also the coefficient of a4b8, since 12C8 = 12C4
13. 13. Example 2. In the expansion of (x − y)15, calculate the coefficients of x3y12and x2y13.15· 14· 131· 2· 3= 455.The coefficient of x2y13, on the other hand, is negative, because the exponent of y isodd. The coefficient is − 15C13= − 15C2. We have−15· 141· 2= −105Example 3. Write the 5th term in the expansion of (a + b)10.Solution. In the 1st term, k = 0. In the 2nd term, k = 1. And so on. The index k ofeach term is one less than the ordinal number of the term. Thus in the 5th term, k = 4.The exponent of b is 4. The 5th term is10C4a6b4=10· 9· 8· 71· 2· 3· 4a6b4= 210 a6b4Solution. The coefficient of x3y12 is positive, because the exponent of y is even.That coefficient is 15C12. But 15C12 = 15C3, and so we have
14. 14. Binomial Theorem for positive integral indexAny expression containing two terms only is called binomial expression eg. a+b, 1 + ab etcFor positive integer n( )n n n 0 n n–1 1 n n–2 2 n 1 n–1 n 0 n0 1 2 n–1 na b c a b c a b c a b ... c a b c a b+ = + + + + +nn n–r rrr 0c a b== ∑where( ) ( )n nr n–rn! n!c c for 0 r nr! n – r ! n – r !r!= = = ≤ ≤ are called binomial coefficients( ) ( )nrn n – 1 ... n – r 1C ,1.2.3...r+= numerator contains r factors10 10 107 10–7 310! 10.9.8C 120 C C7! 3! 3.2.1= = = = =Binomial theorem
15. 15. Special cases of binomial theorem( ) ( )n nn n n n–1 n n–2 2 n n0 1 2 nx – y c x – c x y c x y ... –1 c y= + +( )nr n n–r rrr 0–1 c x y== ∑( )nn n n n 2 n n n r0 1 2 n rr 01 x c c x c x ... c x c x=+ = + + + + = ∑in ascending powers of x( )n n n n n–1 n0 1 n1 x c x c x ... c+ = + + +nn n–rrr 0c x== ∑ ( )nx 1= +in descending powers of x
16. 16. Solution :Expand (x + y)4+(x - y)4and hencefind the value of ( ) ( )4 42 1 2 1+ + −( )4 4 4 0 4 3 1 4 2 2 4 1 3 4 0 40 1 2 3 4x y C x y C x y C x y C x y C x y+ = + + + +4 3 2 2 3 4x 4x y 6x y 4xy y= + + + +Similarly ( )4 4 3 2 2 3 4x y x 4x y 6x y 4xy y− = − + − +( ) ( ) ( )4 4 4 2 2 4x y x y 2 x 6x y y∴ + + − = + +( ) ( )4 4 4 2 2 4Hence 2 1 2 1 2 2 6 2 1 1 + + − = + + ÷ =34Illustrative ExamplesIllustrative Examples
17. 17. General term of (a + b)n:n n r rr 1 rT c a b ,r 0,1,2,....,n−+ = =n n 01 0r 0, First Term T c a b= =n n 1 12 1r 1, Second Term T c a b−= =n n r 1 r 1r r 1T c a b ,r 1,2,3,....,n− + −−= =1 2 3 4 5 n n 1r 0 1 2 3 4 n 1 nT T T T T T T += −n+1 termskth term from end is (n-k+2)th term from beginningIllustrative ExamplesIllustrative Examples
18. 18. Solution : 9 r r9r 1 r4x 5T C5 2x−+−   =  ÷  ÷   4 5 4 596 5 1 5 4 54x 5 9! 4 5T T C5 2x 4!5! 5 2 x+−   = = = − ÷  ÷   39.8.7.6 2 .54.3.2.1 x= − 5040x= −Illustrative ExamplesIllustrative ExamplesFind the 6th term in theexpansion ofand its 4th term from the end.952x − ÷ 4x5
19. 19. 9 r r9r 1 r4x 5T C5 2x−+−   =  ÷  ÷   4th term from end = 9-4+2 = 7th term from beginning i.e. T73 6 3 697 6 1 6 3 6 34x 5 9! 4 5T T C5 2x 3!6! 5 2 x+−   = = = ÷  ÷   339.8.7 53.2.1 x=310500x=Illustrative ExamplesIllustrative Examples
20. 20. Middle termCase I: n is even, i.e. number of terms odd only one middle termthn 2term2+  ÷ n nn 2 2n 2 n n12 2 2T T c a b++= =Case II: n is odd, i.e. number of terms even, two middle termsthn 1term2+  ÷ n 1 n 1n 2 2n 1 n 1 n 112 2 2T T c a b+ −+ − −+= =Middle term= ?2n1xx + ÷ thn 3term2+  ÷ n 1 n 1n 2 2n 3 n 1 n 112 2 2T T c a b− ++ + ++= =Illustrative ExamplesIllustrative Examples
21. 21. Greatest CoefficientCaseI: n evenn12nterm T is max i.e. for r2+=Coefficient of middlenn2CCaseII: n oddn 1 n 32 2n 1 n 1term T or T is max i.e. for r or2 2+ +− +=Coefficient of middlen nn 1 n 12 2C or C− +Illustrative ExamplesIllustrative Examples
22. 22. Find the middle term(s) in theexpansion ofand hence find greatest coefficientin the expansion73x3x6 − ÷ ÷ Solution :Number of terms is 7 + 1 = 8 hence 2 middle terms, (7+1)/2 = 4th and (7+3)/2 = 5th( )33 4 13474 3 1 3 3x 7! 3 xT T C 3x6 4!3! 6+ −= = = − ÷ ÷ 131337.6.5 3x 105x3.2.1 82= − = −Illustrative ExamplesIllustrative Examples
23. 23. Find the middle term(s) in the expansionof and hence find greatestcoefficient in the expansion73x3x6 − ÷ ÷ ( )43 3 15375 4 1 4 4x 7! 3 xT T C 3x6 3!4! 6+ −= = = ÷ ÷ Solution :151547.6.5 x 35x3.2.1 482 3= =Hence Greatest coefficient is7 74 37! 7.6.5C or C or 353!4! 3.2.1= =Illustrative ExamplesIllustrative Examples
24. 24. Find the coefficient of x5in the expansionof and term independent of x102313x2x − ÷ Solution :( )r10 r10 2r 1 r 31T C 3x2x−+ = − ÷ r10 10 r 20 2r 3rr1C 3 x2− − − = − ÷ For coefficient of x5 , 20 - 5r = 5 ⇒ r = 3310 10 3 53 1 31T C 3 x2−+ = − ÷ Coefficient of x5= -32805Illustrative ExamplesIllustrative Examples
25. 25. Solution Cont.( )r10 r10 2r 1 r 31T C 3x2x−+ = − ÷ r10 10 r 20 2r 3rr1C 3 x2− − − = − ÷ For term independent of x i.e. coefficient of x0 , 20 - 5r = 0 ⇒ r = 4410 10 44 1 41T C 32−+ = − ÷ Term independent of x765458=Illustrative ExamplesIllustrative Examples
26. 26. Find the term independent ofx in the expansion of81 –13 51x x .2  +  Solution :8 r r1 18 3 5r 1 r1T C x . x2−−+    ÷  ÷= ÷  ÷   8 r r8 r8 3 5r1C x2−− − =  ÷ 40 5r 3r 40 8r8 r 8 r8 815 15r r1 1C x C x2 2− − −− −   = = ÷  ÷   For the term to be independent of x40 8r0 r 515−= ⇒ =Hence sixth term is independent of x and is given by386 51 8! 1T C . 72 5! 3! 8 = = = ÷ 6T 7=Illustrative ExamplesIllustrative Examples
27. 27. Find (i) the coefficient of x9(ii) the termindependent of x, in the expansionof92 1x3x − ÷ Solution :( )r9 r9 2r 1 r1T C x3x−+− =  ÷ r9 18 2r rr1C x3− −− =  ÷ r9 18 3rr1C x3−− =  ÷ i) For Coefficient of x9 , 18-3r = 9 ⇒ r = 33 39 9 9 94 r1 9! 1 28T C x x x3 3! 6! 3 9− − −   = = = ÷  ÷   hence coefficient of x9 is -28/9ii) Term independent of x or coefficientof x0, 18 – 3r = 0 ⇒ r = 66 697 61 9! 1 28T C3 6! 3! 3 243− −   = = = ÷  ÷   Illustrative ExamplesIllustrative Examples