Question

If (1 + x)ⁿ = C₀ + C₁x + C₂x² + ..... + C_nxⁿ, then C₀C₁ + C₁ C₂ + .... + C_n–1C_n is equal to

(a) \(\frac{(2n)!}{(n-1)!(n+1)!}\) 

(b) \(\frac{(2n-1)!}{(n-1)!(n+1)!}\)

(c)  \(\frac{(2n)!}{(n+2)!(n+1)!}\)

(d) None of these

Answer 1

Answer : (a) \(\frac{(2n)!}{(n-1)!(n+1)!}\) 

Given, (1 + x)ⁿ = C₀ + C₁x + C₂ x² + ..... + C_n xⁿ ...(i) 

Also,(x + 1)ⁿ = C₀ xⁿ + C₁x^n–1 + C₂ x^n–2 + ..... + C_n–1 x + C_n ...(ii)

Multiplying (i) by (ii) we get

(1 + x)²ⁿ = (C₀ + C₁x + C₂ x² + ..... + C_n–1 x^n–1 + C_n xⁿ) × (C₀ xⁿ + C₁x^n–1 + C₂ x^n–2 + ..... + C_n–1 x + C_n)

Now equating the coefficient of x^n–1 on both the sides, we get

C₀C₁ + C₁C₂ + ..... + C_n–1C_n = ²ⁿC_n–1

₌ \(\frac{(2n)!}{(n-1)!(n+1)!}\)