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 C₀C_r + C₁C_r+1 + C₂C_r+2 + . . . + C_n-r C_n

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

Solution: We have,

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

Also, C₀xⁿ + C₁x^n-1 + C₂x^n-2 + . . . + C_n = (x + 1)ⁿ …. (2)

Multiplying (1) and (2), we get

(C₀ + C₁x + C₂x² + …. + C_nxⁿ)(C₀xⁿ + C₁x^n-1 + C₂x^n-2 + …+ C_n) = (1 + x)²ⁿ …(3)

Equating coefficient of x^n-r from both sides of (3), we get

C₀C_r + C₁C_r+1 + C₂C_r+2 + … + C_n-rC_n = ²ⁿ C_n-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 : ²ⁿC₀² – ²ⁿC₂² + ²ⁿC₄² – ……..+ (-1)²ⁿ²ⁿC_2n² = (-1)ⁿ ²ⁿC_n

Solution: We have

(1 – x)²ⁿ = ²ⁿC₀ – ²ⁿC₁ x + ²ⁿC₂ x² – . . . . + (-1 )²ⁿ. ²ⁿC_2nx²ⁿ ….(1)

and also ( x + 1)²ⁿ = ²ⁿC₀ x²ⁿ+ ²ⁿC₁ x^2n-1 +. . . + ²ⁿC_2n ….(2)

multiplying (1) and (2), we get

(²ⁿC₀ – ²ⁿC₁ x+ ²ⁿC₂ x² +….+(-1)²ⁿ. ²ⁿC_2nx²ⁿ)(²ⁿC₀ x²ⁿ+²ⁿC₁ x^2n-1+…+²ⁿC_2n)=(1–x²)²ⁿ

equating the coefficient of x²ⁿ, we get

²ⁿC₀² – ²ⁿC₂² + ²ⁿC₄² – ……..+ (-1)²ⁿ²ⁿC_2n² = (-1)ⁿ ²ⁿC_n

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

then prove that :

C₁² +2. C₂² + 3. C₃² + . . . + n. C_n² = (2n−1)! / [(n−1)!]²

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

Differentiating both the sides with respect to x , we get

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

Also, we have

(1+ x)ⁿ = C₀xⁿ + C₁x^n-1 +C₂ x^n-2+ . . . + C_n …. (3)

Multiplying (2) and (3), we get

(C₁ +2C₂ x + 3C₃x² + . . . +n C_nx^n-1)(C₀xⁿ + C₁x^n-1 +C₂ x^n-2+ . . .+ C_n) = n(1+x)^2n-1

Equating the coefficients of x^n-1, we get

C₁² +2. C₂² + 3. C₃² + . . . + n. C_n² = n.^2n-1C_n-1 = (2n−1)! / [(n−1)!]²