Question

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

Answer 1

$=n{2}^{n-2}[2+n-1]=n(n+1)2^{n-2}$
keeping in view $2^3{C}_2$ or $3^3{C}_3$
Differentiate and multiply by x
$n(1+x{)}^{n-1}x=C_{1}x+2C_2{x}^2+3C_3{x}^3+......$
Again differentiate and multiply by x
$nx\left\{\right(n-1\left)\right(1+x{)}^{n-2}\cdot x+1(1+x{)}^{n-1}\}$
$=C_{1}x+2^2{C}_2{x}^2+3^2{C}_3{x}^3+....$
or $n(n-1)(1+x{)}^{n-2}{x}^2+nx(1+x{)}^{n-1}=C_{1}x+2^2{C}_2{x}^2+3^2{C}_3{x}^3+..........$
Differentiate again w.r.t. x.
$n(n-1)\left\{\right(n-2)\cdot (1+x{)}^{n-3}\cdot x^2+(1+x{)}^{n-2}\cdot 2x\}$ $+n\left\{\right(n-1\left)x\right(1+x{)}^{n-2}+1(1+x{)}^{n-1}\}$
$=C_1+2^3{C}_{2}x+3^3{C}_3{x}^2+.....$
Now put x = 1
$n(n-1)\left\{\right(n-2)2^{n-3}+2^{n-2}\cdot 2\}+n(n-1)2^{n-2}+n\cdot 2^{n-1}$
$2^{n-3}\left\{n\right(n-1\left)\right(n-2)+4n(n-1)+2n(n-1)+4n\}$
$=2^{n-3}n(n^2-3n+2+4n-4+2n-2+4)$
$=2^{n-3}n(n^2+3n)=n^2(n+3)2^{n-3}$