4.22 | Binomial Theorem

PROBLEM-SOLVING TACTICS

Summation of series involving binomial coefficients n n n n2 nn nn n n For ()1+ x = C01 + C x + C 2 x ++ ...... C n x , the binomial coefficients are C01 , C , C 2 ,..., C n. A number of series may be formed with these coefficients figuring in the terms of a series. Some standard series of the binomial coefficients are as follows:

nnn n n (a) By putting x = 1, we get C012+ C + C ++ ...... C n = 2 …(i)

nnnn n (b) By putting x =-1, we get C012− C + C − ...... +−() 1 . C n = 0 ...(ii)

n n n n1− (c) On adding (i) and (ii), we get C024+ C + C += ...... 2 …(iii)

n n n n1− (d) On subtracting (ii) from (i), we get C135+ C + C += ...... 2 …(iv)

2n 2n 2n 2n 2n 2n− 1 (e) C012+ C + C ++ ...... C n1n− + C = 2

2n 2n 2n 2n 2n 2n 2n Proof: From the expansion of ()1x+ , we get C0+ C 1 + C 2 ++ ...... C 2n− 1 + C 2n = 2

⇒2n + 2n + 2n ++ 2n + 2n = 2n 2n= 2n 2n = 2n 2() C0 C 1 C 2 ...... C n1− C n 2 [ C0 C, 2n C 1 C 2n− 1 and so on. ]

2n+++ 1 2n 1 2n 1 2n + 1 2n (f) C012+ C + C ++ ...... C n = 2

Proof: (as above)

nn n nn n n–1 (g) Sum of the first half of C01+ C ++ ... C n= Sum of the last half of C01+ C ++ ... C n= 2

n n n2 nn n (h) Bino-geometric series: C01+ C x + C 2 x ++ ...... C n x =+() 1 x

nn n n (i) Bino-arithmetic series: a C01++()()() a d C ++ a 2d C 2 + ...... ++ a nd C n Consider an AP -a, (a+d), (a+2d), … , (a+nd) nnn n Sequence of Binomial Co-efficient - C012 , C , C ,...... , C n A bino-arithmetic series is nothing but the sum of the products of corresponding terms of the sequences. It can be added in two ways. (i) By elimination of r in the multiplier of binomial coefficient from the (r+1)th term of the series

n= n1− ()By using r. Cr n C r1− n (ii) By differentiating the expansion of x1xdd()+ . Mathematics | 4.23

nC nnCC n C (j) Bino-harmonic series: 0 +12 + ++...... n a ad++ a2d and + 11 1 1 Consider an HP - , , ,..., aada2dand++ +

nnn n Sequence of Binomial Co-efficient - C012 , C , C ,...... , C n

It is obtained by the sum of the products of corresponding terms of the sequences. Such series are calculated in two ways : (i) By elimination of r in the multiplier of binomial coefficient from the (r th+ 1) term of the series

11n n1+ By using Cr= Cr1+ r1++ n1

(ii) By integrating suitable expansion. For explanation see illustration 2

nn nn nn n n (k) Bino-binomial series: C0r . C+ C 1r12r2 . C++ + C . C ++ ...... C nrr − . C

m n mn m n mn or, C0r . C+ C 1r12r2 . C−− + C . C ++ ..... C r0 . C

As the name suggests such series are obtained by multiplying two binomial expansion, one involving the first factors as coefficient and the other involving the second factors as coefficient. They can be calculated by equating coefficients of a suitable power on both sides. For explanation see illustration 4

FORMULAE SHEET

Binomial theorem for any positive integral index:

n n n n n n1−− n n2 2 n nr − r n n n nr− r (x+= a) C01 x + C x a + C 2 x a + ...... + Cr x a ++ ...... Cn a =∑ C r x a r0= n nr− r th (a) General term – Tr1+ = Cx r ais the (r + 1) term from beginning. th th (b) (m + 1) term from the end = (n – m + 1) from beginning = Tn–m+1 (c) Middle term th n (i) If n is even then middle term = + 1 term 2

th th n1+ n3+ (ii) If n is odd then middle term = and  2 2 Binomial coefficient of middle term is the greatest binomial coefficient.

To determine a particular term in the given expansion:

th α 1 n th term Let the given expansion be x ± , if x occurs in Tr+1 (r + 1) then r is given by nrα−( α+β) = m and for xβ x0 ,nα− r( α+β) = 0 4.24 | Binomial Theorem

Properties of Binomial coefficients:

nnn n n For the sake of convenience the coefficients C012 , C , C ...... C r ...... C n are usually denoted by C01 ,C ,...... C r ...... C n respectively.

n C012+++ C C ...... += C n 2

C0123−+−+ C C C ...... += C n 0

n1− C024+++ C C ...... =+++ C 135 C C ...... = 2

n nn1− nCC= n1−− = ⋅ n2C and so on...... rr r1−− rr1− r2

2n! 2nC = nr+ (n−+ r!n) ( r!)

n n n1+ CCr+= r1− C r

n n1− C123++++ 2C 3C ...... C n = n.2

C123−+ 2C 3C ...... = 0

n1− C012+ 2C + 3C + ...... ++( n 1) Cn =+( n 2) 2 (2n) ! C2+++ C 2 C 2 ...... += C2 =2nC 012 n 2 n (n!)  2222  0, if n is odd C−+−+= C C C .....  0123 − n/2 n ( 1) Cn/2 , if n is even

2n++ 1 2n 1 2n + 1 2n + 1 2n + 1 2n + 1 2n Note: C0+ C 1 + ...... + Cn = C n1++ + C n2 ++ ...... C2n1 + = 2

n1+ CC C 21− −1Cn 12 n CC12C3 ( ) n 1 C0 ++++...... = ; C −+−...... + = 2 3 n1++ n1 0 2 3 4 n1++ n1 (a) Greatest term:

(n+ 1a) (i) If ∈ Z (integer) then the expansion has two greatest terms. These are kth and (k + 1)th where x xa+ and a are +ve real numbers.

(n+ 1a) (n+ 1a) (ii) If ∉ Z then the expansion has only one greatest term. This is (K + 1)th term k =  + + xa xa denotes greatest integer less than or equal to x}

(b) Multinomial theorem:

n n! rr r Generalized (x++ x ...... + x ) = x12 x .....x k 1 2 k ∑ r !r !...... r ! 12 k r12++ r .....r k = n 12 k

m mn1+− (c) Total no. of terms in the expansion (x12++ x ...... x n) is Cn1−