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https://docs.google.com/document/d/1VGhpH0_K8gkyWbEIdsi_46WujNmug7Te/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true Binomial Theorem and its Application Binomial theorem for a positive integral index, General term and middle term, Properties of Binomial coefficients and Simple applications C H A P T E R BINOMIAL EXPRESSION An algebraic expression containing only two terms is called a binomial CHAPTER INCLUDES : expression. Such as a + x, 2x + 5y, expressions. 2x + 1 , y a + b x etc. all are binomal Binomial expression, Binomial thorem for positive integral index BINOMIAL THEOREM FOR POSITIVE INTEGRAL INDEX If x, y ∈ R, then ∀ n ∈ N (x + y)n = nC xn + nC xn – 1y + nC xn – 2y2 + .... + nC xn – ry r + .... + nC yn Here all nC 's are called binomial coefficient General term in the expansion of ( x + y)n Middle term in the expansion of (x + y) n pth term from the end in expansion of (x + y)n Numerically greatest term in the expansion of (1 + x) n Properties of binomial coefficients Some relation involving binomial coefficients Some useful expansions. Solved examples In the expansion of (x + y)n, if the sum of odd terms be A and that of even terms be B, then 4AB equals. (1) (x + y)n + (x – y)n (2) (x + y)n – (x – y)n (3) (x + y)2n – (x – y)2n (4) (x + y)2n + (x – y)2n Solution : (x + y)n = A + B (x – y)n = A – B ∴ 4AB = (A + B)2 – (A – B)2 = (x + y)2n – (x – y)2n Hence Ans is (3) General term in the expansion of (x + y)n The (r + 1)th term = T is called the general term. Find (i) coefficient of x9 and (ii) term independent of x in the expansion of ⎛ x 2 − ⎝ 1 ⎞9 ⎟ 3x ⎠ Solution : ⎛ 1 ⎞9 Tr +1 = nCr ⋅ (first term)n−r ⋅ (second term)r in the expansion of ⎜ x 2 − ⎝ ⎟ 3x ⎠ ∴ T = 9C (x 2 )9−r ⋅ (−1)r ⋅ ⎛ 1 ⎞r ⎟ = 9Cr ⋅ (−1)r ⋅ x18−3r ⎝ 3x ⎠ 3r We want coefficient of x9 ∴ 18 – 3r = 9 3r = 9 ; r = 3 ∴ Coefficient of x9 is 9C3 ⋅ (−1)3 ⋅ 1 33 = − 28 9 For the term independent of x, we want the power of x to be 0. ∴ 18 – 3r = 0 ; r = 6 T6+1 = 9C6 ⋅ (−1)6 36 = 9 × 8 × 7 6 × 36 = 28 . 243 Middle term in the expansion of (x + y)n Case 1. n = 2m ; m ∈ N Then, number of terms in the expansion = n + 1 = 2m + 1; Thus middle term is Tm + 1 ⎛ n ⎞th i.e. ⎜ 2 + 1⎟ term ⎝ ⎠ n n Tm +1 = nCn 2 x 2 , y 2 Case 2. n = 2m + 1 ; m ∈ N Then, number of terms in the expansion is = n + 1 = 2m + 2. There are two middle terms, viz.Tm +1 and Tm +2 . ⎛ n + 1⎞th ⎛ n + 3 ⎞th i.e. ⎜ ⎟ ⎝ 2 ⎠ and ⎜ ⎟ ⎝ 2 ⎠ term Tm +1 = nCn −1 ⋅ x 2 n +1 2 n −1 ⋅ y 2 Tm +2 = nCn+1 ⋅ x 2 n −1 2 n +1 ⋅ y 2 Find the middle term in the expansion of (i) (3x + 2y)10 (ii) ⎛ 2x + ⎝ 3 ⎞9 ⎟ x ⎠ Solution : ⎛ 10 ⎞th Since 10 is even number hence middle term will be ⎜ ⎝ + 1⎟ ⎠ i.e. 6th term T = 10C (3x)10 – 5 (2y)5 = 10.9.8.7.6 × 35 × 25 x 5 y 5 5.4.3.2.1 = 7 × 27 × 37 × x5y5 ⎛ 9 + 1⎞th ⎛ 9 + 3 ⎞th Since 9 is odd number, so there will be two middle term i.e. ⎜ ⎟ ⎝ 2 ⎠ and ⎜ ⎝ ⎟ terms ⎠ i.e. 5th and 6th terms ⎛ 3 ⎞4 T = 9C (2x)5 ⎜ ⎟ = 9C .25.34.x 5 4 ⎝ x ⎠ 4 ⎛ 3 ⎞5 T = 9C (2x)4 ⎜ ⎟ = 9C .24.35.x–1 6 5 ⎝ x ⎠ 5 pth term from the end in the expansion of (x + y)n pth term from end = (n + 2 – p)th term from begining = n Cn −p +1x p − 1y (n − p +1) ⎛ x 3 2 ⎞9 Find the 4th term from end in the expansion of ⎜ − 2 ⎟ . ⎜ ⎟ ⎝ ⎠ Solution : 4th term from end = (9 – 4 + 2) = 7th term from begining ⎛ x 3 ⎞3 ⎛ − 2 ⎞6 672 i.e. T7 = 9C6 ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎝ x 2 ⎠ x 3 Numerically greatest term in (1 + x)n The method for finding the greatest term is as below : Find the value of K = (n + 1) | x | . 1 + | x | If K is an integer, then TK and TK + 1 both are equal and greatest. If K is not an integer then T[K] + 1 is the greatest term where [K] is the greatest integral part of K. ⎛ y ⎞n ⎛ y ⎞n To find the greatest term in (x + y)n = xn ⎜1 + ⎟ x , find the greatest term in ⎜1 + ⎟ and multiply with xn. ⎝ ⎠ ⎝ x ⎠ Find the greatest term in the expansion of (2 + 3x)9 if x = 3 . 2 Solution : In the expansion of (x + a)n ≥ 1 ⇒ r ≤ n + 1 ∴ r ≤ 9 + 1 = 10 × 9 = 90 = 6 + 12 1+ 2 3 ⋅ 3 2 9 + 4 13 13 ∴ rmax. = 6 ⎛ 3 ⎞6 9 ! 312 21 Tr +1 = 9C6 ⋅ 29−6 ⋅ ⎜3 ⋅ ⎟ ⎝ 2 ⎠ = 6 ! 3 ! ⋅ 8 ⋅ 26 = ⋅ 312 . 2 Properties of Binomial Coefficients In the expansion of (1 + x)n, the coefficients, nC0, nC1, nC2 ..... nCn are denoted by C0, C1, C2 Cn then. If n is even, then greatest coefficient is nCn/2. If n is odd, then greatest coefficients are nC(n –1)/2 or nC(n +1)/2 3. r.nCr = n.n – 1Cr – 1 0 < r ≤ n. n C 4. r = r 1 n + 1Cr + 1 n + 1 nC = n n − 1C r r (r − 1) n r n r − 1 = n − r + 1 r n + n r − 1 r n + 1 Cr nCx = n y ⇒ x = y or x + y = n nCr = nCn – r Some Relation involving Binomial Coefficients (1 + x)n = 1 + nC x + nC x2 + ......... + nC xn …(1) Relation (1) is important. Many properties of binomial coefficients are obtained by Giving different values to x in (1) and/or Differentiating (1) with respect to x and then putting convenient values of x, and/or Integrating (1) with respect to x and then putting convenient values of x. Some important relation are given below : 1. ∑ n Cr r = 0 = C0 + C1 + C2 + ....... Cn = 2n 2. C + C + C ......... = C + C + = 2n – 1 3. ∑r. nCr r = 1 or 0 = C1 + 2C2 + 3C3 ........ + nCn = n.2n − 1 4. ∑ r = 0 n r r + 1 = 2n + 1 − 1 n + 1 5. C0 – C1 + C2 – C3 + + (–1)n Cn = 0 6. 0 C 2 + .........C 2 = ∑(n r = 0 Cr )2 = 2n Cn = 2n! (n!)2 7. C 2 − C 2 + C 2 − C 2 + (−1) nC 2 = ⎡0, if n is odd 0 1 2 3 n ⎢ ⎢⎣(–1) n/2.nCn / 2 if n is even 8. C C + C C + .......... + C C = 2nC Show that C 2 + 2C 2 +.......+(n + 1)C 2 = (n + 2) (2n – 1) 0 1 n n! (n – 1)! Solution : Let S = C 2 + 2C 2 + +(n + 1)C 2 Also S = (n + 1)C 2 + nC 2 + (n – 1)C2 + + C 2 S + S = (n + 2)C 2 + (n + 2) C 2 + +(n + 2)C 2 2S = (n + 2)[C 2 + C 2 +......+C 2] = (n + 2) (2n) ! 0 1 n S = (n + 2)(2n − 1) ! n !(n − 1) ! (n !)2 SOLVED EXAMPLES Example 1 : If the coefficients of rth, (r + 1)th and (r + 2)th terms in the expansion of (1 + x)14 are in A.P., find r. Solution : In the expansion of (1 + x)14, the rth, (r + 1)th and (r + 2)th terms are Tr = 14Cr −1xr −1 ; Tr +1 = 14Cr xr ; Tr +2 = 14Cr +1 xr +1 and the coefficients are 14Cr −1, 14Cr and 14Cr +1 respectively ∴ 2 ⋅ (14Cr ) = 14Cr −1 + 14Cr +1 ∴ 2 ⋅ 14 ! r ! (14 − r ) ! = 14 ! + (r − 1) ! (15 − r ) ! 14 ! (r + 1) ! (13 − r ) ! 2 ∴ r (14 − r ) = 1 + (15 − r ) (14 − r ) 1 (r + 1) ⋅ r Multiplying throughout by r(r + 1) (14 – r) (15 – r), we have 2(−r 2 + 14r + 15) = r 2 + r + 210 − 29r + r 2 r 2 − 14r + 45 = 0 ; r = 5 or 9 Example 2 : Evaluate (0.99)15 correct to four decimal places. Solution : (0.99)15 = (1 – 0.01)15 = 1− 15C1(0.01) + 15C2(0.01)2 − 15C3 (0.01)3 = 1− 15(0.01) + 105(0.01)2 − 455 (0.01)3 = 1− 0.15 + 0.0105 − 0.000455 ∴ (0.99)15 ≅ 1.0105 − 0.1504 = 0.8601 Example 3 : Find the coefficient of x4 in the expansion of (1+ x + x 2 + x 3 )11 Solution : 1+ x a (1+ x + x 2 + x 3 )11 = (1+ x)11 ⋅ (1+ x 2 )11 Terms in the expansion of (1 + x2)11 would involve even powers of x only. ∴ Term involving x4 in (1 + x)11·(1 + x2)11 can arise in following ways : x4 from (1 + x)11 and x0 from (1 + x2)11 x2 from (1 + x)11 and x2 from (1 + x2)11 x0 from (1 + x)11 and x4 from (1 + x2)11 ∴ Coefficient of x4 in (1+ x + x 2 + x 3 )11 is 11C4 ⋅ 11C0 + 11C2 ⋅ 11C1 + 11C0 ⋅ 11C2 = 11×10 × 9 × 8 + 11× 10 ⋅11+ 11×10 = 330 + 605 + 55 = 990. 24 2 2 Example 4 : Show that 6n + 2 + 72n + 1 is divisible by 43 ∀ n ∈ N . Solution : 6n +2 + 72n+1 = 36 ⋅ 6n + 7 ⋅ (49)n = (43 − 7) ⋅ 6n + 7 ⋅ (43 + 6)n = 43 ⋅ 6n − 7 ⋅ 6n + 7 ⋅ (43k + 6n ) (write the binomial expansion of (43 + 6)n) ≡ multiple of 43 Example 5 : Prove the following : (i) C0Cr C1C r +1 C2C r +2 + + C n −r ⋅ Cn = (2n) ! (n − r ) ! (n + r ) ! In particular, C 2 + C 2 + ......... + C 2 = (2n) ! 0 1 n (n ! )2 (ii) ∑∑ 0 ≤ i < j ≤ n Ci C j = 2n −1 − (2n) ! . 2 ⋅ n ! ⋅ n ! Solution : (i) (1+ x )2n 2n = j =0 2nC j x j …(1) (1+ x )2n = (1+ x)n ⋅ (x + 1)n = (1+ nC1x + nC2 x 2 + ..... + nCn xn ) ( nC0 xn + nC1xn−1 + + nCn ) …(2) Coefficient of xn + r in (1) is 2nCn + r Coefficient of xn + r in (2) is C0 ⋅Cr + C1 ⋅Cr +1 + C2 ⋅Cr +2 + + Cn −r ⋅Cn ∴ C0Cr C1C r +1 + + C n−r ⋅ Cn = (2n) ! . (n − r ) ! (n + r ) ! (ii) C0 + C1 + ....... + Cn = 2n where Cr = nCr ∴ (C0 + C1 + + Cn )2 = 22n n  i =1 2 + 2 ∑ ∑ 0 ≤ i < j ≤ n Ci C j = 22n ⎛ 2n 2n ! ⎞ 1 ∑∑Ci Cj = ⎜ 2 ⎟ (n ! )2 ⎟ 2 0≤i < j ≤n ⎝ ⎠ ❑ ❑ ❑