# Properties of Binomial Coefficient | Solved Examples

Properties of Binomial Coefficient :
For the sake of convenience the coefficients nCo , nC1 , . . . ,nCr , . . . , nCn are usually denoted by Co , C1 , . . . , Cr , . . . ,Cn respectively

∎ Co + C1 + C2 +. . . . . + Cn = 2n

∎ Co – C1 + C2 -. . . . . Cn = 0

∎ Co + C2 + C4 +. . . . . = C1 + C3 + C5 +. . . . . = 2n-1

nCr + nCr-1 = n+1Cr

∎ r nCr = n n-1Cr-1

### Some Important results

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

Differentiating on both sides we have ,

n(1 + x)n-1 = C1 + 2C2x + 3C3x2 + . . . + nCnxn-1 . . . (1)

Putting x = 1 ,

n 2n-1 = C1 + 2C2 + 3C3 + . . . + n Cn

Putting x = −1

0 = C1 – 2C2 + . . . + (−1)n-1 n Cn

Differentiating (1) again and again we will have different results.

(ii) Integrating , (1 + x)n = C0 + C1x + C2x2 + . . . + Cnxn , we have,

$\large \frac{(1+x)^{n+1}}{n+1} + c = C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{3} + … + \frac{C_n x^{n+1}}{n+1}$ (where c is a constant)

Put x = 0, we get $\large c = -\frac{1}{n+1}$

Therefore

$\large \frac{(1+x)^{n+1}-1}{n+1} = C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{3} + … + \frac{C_n x^{n+1}}{n+1}$ …(1)

Put x = 1 in (1) we get

$\large \frac{2^{n+1}-1}{n+1} = C_0 + \frac{C_1}{2} + \frac{C_2 }{3} + … + \frac{C_n}{n+1}$

Put x = -1 in (1) we get,

$\large \frac{1}{n+1} = C_0 – \frac{C_1}{2} + \frac{C_2 }{3} – …..$

(i) Problems Related to series of Binomial coefficients in which each term is a product of an integer and a binomial coefficient i.e. in the form k nCr , k ∈ I

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

C0 + 2C1 + 3C2 + …. + (n + 1)Cn = (2 + n)2n-1

Solution: (1 + x)n = C0 + xC1 + x2C2 + …. + xn Cn

Multiplying with x on both sides, we get

x (1 + x)n = xC0 + x2C1 + x3C2 + … + xn+1 Cn … (1)

Differentiating (1) with respect to x, we get

nx(1 + x)n-1 + (1 + x)n = C0 + 2xC1 + 3x2C2 + …… + (n + 1)xnCn

Put x = 1 ,

=> (2 + n)2n-1 = C0 + 2C1 + 3C2 + ….. + (n + 1) Cn

(ii) Problems related to series of binomial coefficients in which each term is a binomial coefficient divided by an integer i.e. in the form of $\large \frac{n_{C_r}}{k}$; , k ∈ I

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

$\large \frac{C_0}{2} + \frac{C_1}{3} + \frac{C_2}{4} + … + \frac{C_n}{n+2}= \frac{n.2^{n+1} + 1}{(n+1)(n+2)}$

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

⇒  x (1 + x)n = C0x + C1x2 + C2x 3 + ….. + Cnxn+1  …. (1)

Integrating both the sides of (1) with respect to x

$\large \frac{x(1+x)^{n+1}}{n+1} – \int (\frac{(1+x)^{n+1}}{n+1})dx = \frac{x^2}{2}C_0 + \frac{x^3}{3}C_1 + \frac{x^4}{4}C_2 + …+ \frac{x^{n+2}}{n+2}C_n$

$\large \frac{x(1+x)^{n+1}}{n+1} – \frac{(1+x)^{n+2}}{(n+1)(n+2)} + k = \frac{x^2}{2}C_0 + \frac{x^3}{3}C_1 + \frac{x^4}{4}C_2 + …+ \frac{x^{n+2}}{n+2}C_n$

Put x = 0 ,

$\large k = \frac{1}{(n+1)(n+2)}$

$\large \frac{x(1+x)^{n+1}}{n+1} – \frac{(1+x)^{n+2}}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} = \frac{x^2}{2}C_0 + \frac{x^3}{3}C_1 + \frac{x^4}{4}C_2 + …+ \frac{x^{n+2}}{n+2}C_n$

Put x = 1 ,

$\large \frac{n.2^{n+1} + 1}{(n+1)(n+2)} = \frac{C_0}{2} + \frac{C_1}{3} + \frac{C_2}{4} + … + \frac{C_n}{n+2}$