Question

Prove that: $1^2.C_1+2^2.C_2+3^2.C_3+.....n^2.C_n=n(n+1)2^{n-2}$

Answer 1

$C_0+C_{1}x+C_2{x}^2+.....C_n{x}^n=(1+x{)}^n$
$C_2{x}^2$ occurs in $(1+x{)}^n$. We want $2^2{C}_2$ in the L.H.S. Diff to get $2C_{2}X$ but we want $2^2{C}_2>$ Again multiply by x to get $2C_2{X}^2$ . Again diff to get $2^2{C}_{2}X$ and finally put x = 1. The operations done in L.H.S. should also be done in R.H.S. All this is exhibited below.
Differentiating the expansion of $(1+x{)}^n$ we get
$n{(1+x)}^{n-1}=C_1+2C_{2}x+3C_2{x}^2+.......+n{C}_n{x}^{n-1}$ ....(1)
Keeping in view the form of question we multiply both sides of (1) by $x$

$nx{(1+x)}^{n-1}=C_{1}x+2C_2{x}^2+3C_2{x}^3+.......+n{C}_n{x}^n$
Now differentiate w.r.t. x.
$n[1\cdot (1+x{)}^{n-1}+x\cdot (n-1)(1+x{)}^{n-2}]$ .....(2)
$=C_1+2^2{C}_{2}x+3^2{C}_3{x}^2+.......+n^2{C}_2{x}^{n-1}$
Now put x = 1
$n[2^{n-1}+(n-1\left)\right(2^{n-2}\left)\right]=C_1+2^2{C}_2+3^2{C}_3+..........+n^2{C}_n$
$=n{2}^{n-2}[2+n-1]=n(n+1)2^{n-2}$.