Question

Find the sum of $1^3+2^3+\frac{3^3}{\lfloor 2}+\frac{4^3}{\lfloor 3}+\frac{5^3}{\lfloor 4}+....$

Answer 1

$u_{n+1}=\frac{{(n+1)}^3}{n!}=\frac{n(n-1)(n-2)+6n(n-1)+7n+1}{n!}$

$=\frac{1}{(n-3)!}+\frac{6}{(n-2)!}+\frac{7}{(n-1)!}+\frac{1}{n!}$

$\because \sum \frac{1}{r!}=e$

$\therefore S_{\infty }=\sum \frac{1}{(n-3)!}+6\sum \frac{1}{(n-2)!}+7\sum \frac{1}{(n-1)!}+\sum \frac{1}{n!}$

$=e+6e+7e+e$

$=15e$