Question

If n is a positive integer greater than $3$, show that $n^3+\frac{n(n-1)}{\lfloor 2}(n-2{)}^3+\frac{n(n-1)(n-2)(n-3)}{\lfloor 4}(n-4{)}^3+....=n^2(n+3)2^{n-4}$.

Answer 1

$\Rightarrow {(e^x+1)}^n=e^{nx}+c_1{e}^{(n-1)x}+c_2{e}^{(n-2)x}+....$

${(e^x-1)}^n=e^{nx}-c_1{e}^{(n-1)x}+c_2{e}^{(n-2)x}+....$

$2\{e^{nx}+c_2{e}^{(n-2)x}+c_4{e}^{(n-4)x}+....\}$

$={(e^x+1)}^n+{(e^x-1)}^n$

Equating co efficients of $x^3$, we have -

$\frac{2}{3!}\{n^3+\frac{n(n-1)}{2!}{(n-3)}^3+...\}=$ co efficient of $x^3$ in ${(e^x+1)}^n+{(e^x-1)}^n$

$\Rightarrow \frac{2}{6}S=$ co efficient of $x^3$ in ${(2+x+\frac{x^2}{2}+\frac{x^3}{6}+....)}^n$

i.e., in,

$n.2^{n-1}(x+\frac{x^2}{2}+\frac{x^3}{6})+\frac{n(n-1)}{2!}{2}^{n-3}{(x+\frac{x^2}{2})}^2+\frac{n(n-1)(n-2)}{3!}{2}^{n-3}.x^3$

$\therefore \frac{2S}{6}=\frac{n.2^{n-1}}{6}+\frac{n(n-1)}{2!}{2}^{n-2}+\frac{n(n-1)(n-2)}{3!}{2}^{n-3}$

$S=n^2(n+3).2^{n-4}$