Question

The sum of first $n$ terms of the series $1^3+{3.2}^2+3^3+{3.4}^2+5^3+{3.6}^2+....$, where $n$ is even be $\therefore S_n=\frac{n}{k}[n^m+4n^2+10n+p]$. Find $k+m-p$ ?

Answer 1

Case I. $n$ $even=2m$, say
$T_r$ of $1,3,5,...=2r-1$
$T_r$ of $2,4,\mathrm{6...}=2r$
$S_n=S_{2m}=\sum _{r=1}^m{(2r-1)}^3+3\sum _{r=1}^m{\left(2r\right)}^2$
$=\sum _{r=1}^m[8r^3-3{\left(r\right)}^2+3\left(2r\right)-1]+12\sum _{r=1}^m{r}^2$
$=8\sum _{r=1}^m{r}^3+6\sum _{r=1}^m(r^2+r)-\sum _{r=1}^{m}1$
$2m^2{(2+1)}^2+3m(m+1)-m$
$=m[2m^3+4m^2+5m+2]$
Put $2m=2$ or $=m=\frac{n}{2}$
$\therefore S_n=\frac{n}{8}[n^3+4n^2+10n+8]$

$k+m-p=8+3-8=3$