Binomial theorem

Question

If (1+x)ⁿ = C₀​+C₁​x+C₂​x²+....+C_n​xⁿ, then C_o²​+C₁²​+C₂²​+....+C_n²​ is equal to

Answer 1

Correct option is (b) \(\frac{(2n)!}{n!n!}\)

Given (1 + x)ⁿ = C₀ ​+ C₁​x + C₂​x² + ....+ C_n​xⁿ  ...(i)

Clearly, RHS is the term independent of x and hence it is equal to the term independent of x in the product.

\(\frac1{x^n}\)​(1 + x)²ⁿ or term containing xⁿ in (1 + x)²ⁿ

Clearly the coefficient of xⁿ in (1 + x)²ⁿ occurs in T_n+1​ and is equal to ²ⁿC_n​= \(\frac{(2n)!}{n!n!}\)