Question

If $C_0,C_1,C_2...C_n$ denote the Binomial coefficients in the expansion of ${(1+x)}^n$, then the expression
$1^2.C_1+2^2{C}_2+3^2{C}_3+...+n^2{C}_n$ equals

• A. $n(n+1)2^{n-3}$
• B. $n(n+1)2^n$
• C. $n(n+1)2^{n-1}$
• D. $n(n+1)2^{n-2}$

Answer 1

Answer: (D)

$n(n+1)2^{n-2}$

$(1+x{)}^n=n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+....{}^n{C}_n{x}^n$
Differentiating with respect to $x$.
$n(1+x{)}^{n-1}={}^n{C}_1+2{}^n{C}_{2}x+....n{}^n{C}_n{x}^{n-1}$
Multiplying, by $x$ on both sides.
$nx(1+x{)}^{n-1}={}^n{C}_{1}x+2{}^n{C}_2{x}^2+....n{}^n{C}_n{x}^n$
Differentiating with respect to $x$.
$n(1+x{)}^{n-1}+n(n-1\left)x\right(1+x{)}^{n-2}={}^n{C}_1+2^2{}^n{C}_2{x}^1+3^2{}^n{C}_3{x}^2+....n^{2}n{C}_n{x}^{n-1}$
Substituting $x=1$, we get
$n(2{)}^{n-1}+n(n-1)2^{n-2}={}^n{C}_1+2^2{}^n{C}_2+3^2{}^n{C}_3+....n^{2}n{C}_n$
Hence
${}^n{C}_1+2^2{}^n{C}_2+3^2{}^n{C}_3+....n^{2}n{C}_n=n{2}^{n-2}[2+n-1]$
$=n(n+1)2^{n-2}$