# Binomial Coefficients | Solved Problems

(iii) Problems related to series of binomial coefficients in which each term is a product of two binomial coefficients.

Illustration : Prove that C0Cr + C1Cr+1 + C2Cr+2 + . . . + Cn-r Cn

$\large = \frac{(2n)!}{(n-r)!(n+r)!}$

Solution: We have,

C0 + C1x + C2x2 + . . . + Cnxn = (1 + x)n …. (1)

Also, C0xn + C1xn-1 + C2xn-2 + . . . + Cn = (x + 1)n …. (2)

Multiplying (1) and (2), we get

(C0 + C1x + C2x2 + …. + Cnxn)(C0xn + C1xn-1 + C2xn-2 + …+ Cn) = (1 + x)2n …(3)

Equating coefficient of xn-r from both sides of (3), we get

C0Cr + C1Cr+1 + C2Cr+2 + … + Cn-rCn = 2n Cn-r

$\large = \frac{(2n)!}{(n-r)!(n+r)!}$

Illustration : If  $\large (1+x)^n = \Sigma_{r=0}^{n} n_{C_r} x^r$
then  prove  that : 2nC022nC22 + 2nC42 – ……..+ (-1)2n 2nC2n2 = (-1) 2nCn

Solution:       We have

(1 – x)2n = 2nC02nC1 x + 2nC2 x2 – . . . . + (-1 )2n. 2nC2nx2n      ….(1)

and  also ( x + 1)2n = 2nC0 x2n + 2nC1 x2n-1 +. . . + 2nC2n            ….(2)

multiplying  (1) and (2),  we get

(2nC02nC1 x+ 2nC2 x2 +….+(-1)2n. 2nC2nx2n)(2nC0 x2n+2nC1 x2n-1+…+2nC2n)=(1–x2)2n

equating   the   coefficient   of  x2n, we get

2nC022nC22 + 2nC42 – ……..+ (-1)2n 2nC2n2 = (-1) 2nCn

Illustration : If  $\large (1+x)^n = \Sigma_{r=0}^{n} n_{C_r} x^r$

then prove that :

C12 +2. C22 + 3. C32 + . . . + n. Cn2 = (2n−1)! / [(n−1)!]2

Solution: ( 1+ x)n = C0 + C1x +C2 x2 +C2x3 + . . .+ nCnxn …. (1)

Differentiating both the sides with respect to x , we get

n( 1+ x)n-1 = C1 +2C2 x + 3C2x2 + . . . + nCnxn-1 …. (2)

Also, we have

(1+ x)n = C0xn + C1xn-1 +C2 xn-2 + . . . + Cn …. (3)

Multiplying (2) and (3), we get

(C1 +2C2 x + 3C3x2 + . . . +n Cnxn-1)(C0xn + C1xn-1 +C2 xn-2 + . . .+ Cn) = n(1+x)2n-1

Equating the coefficients of xn-1, we get

C12 +2. C22 + 3. C32 + . . . + n. Cn2 = n.2n-1Cn-1 = (2n−1)! / [(n−1)!]2