Properties of Binomial Coefficient | Solved Examples

Properties of Binomial Coefficient :
For the sake of convenience the coefficients ⁿC_o , ⁿC₁ , . . . ,ⁿC_r , . . . , ⁿC_n are usually denoted by C_o , C₁ , . . . , C_r , . . . ,C_n respectively

∎ C_o + C₁ + C₂ +. . . . . + C_n = 2ⁿ

∎ C_o – C₁ + C₂ -. . . . . C_n = 0

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

∎ ⁿCr + ⁿC_r-1 = ⁿ⁺¹C_r

∎ r ⁿC_r = n ^n-1C_r-1

Some Important results

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

Differentiating on both sides we have ,

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

Putting x = 1 ,

n 2^n-1 = C₁ + 2C₂ + 3C₃ + . . . + n C_n

Putting x = −1

0 = C₁ – 2C₂ + . . . + (−1)^n-1 n C_n

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

(ii) Integrating , (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . + C_nxⁿ , 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 ⁿC_r , k ∈ I

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

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

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

Multiplying with x on both sides, we get

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

Differentiating (1) with respect to x, we get

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

Put x = 1 ,

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

(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 + ….. + C_nxⁿ

⇒  x (1 + x)ⁿ = C₀x + C₁x² + C₂x ³ + ….. + C_nxⁿ⁺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} $

Also Read :

→ Binomial Theorem : General Term in the Expansion → Middle Terms in Binomial Expansion → Problems Related to Binomial coefficients → Binomial Theorem for any index

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