If the sum of the n terms of the series  $1^3+{3.2}^2+3^3+{3.4}^2+5^3+{3.6}^2+\mathrm{.....},$  where n is an even number, is given by  $\frac{n}{k}(n^3+a{n}^2+bn+c)$ then $b-a+c-k$  is

We have 
 $S=1^3+{3.2}^2+3^3+{3.4}^2+5^3+{3.6}^2+\mathrm{...}$
 $S=(1^3+3^3+5^3+\mathrm{....})+3.(2^2+4^2+6^2+\mathrm{...})$
 $S=(1^3+3^3+5^3+\mathrm{....})+12(1^2+2^2+3^2+\mathrm{...})$
Where  $S_1=1^3+3^3+5^3+\mathrm{.....}and{S}_2=1^2+2^2+3^2+\mathrm{...}$
Now case arise 
When n is , say even ,say n=2m  $m\in N$
In this case $S_1,and{S}_2$  both contain m terms 
 $\therefore S_1=1^3+3^3+5^3+\mathrm{...}+{(2m-1)}^3$
 $\sum _{r=1}^M{(2r-1)}^3$
$\sum _{r=1}^m{(8r^3-12r^2+6r-1)}^{ }$ 
 $=8\sum _{r=1}^m{r}^3-12\sum _{r=1}^m{r}^2+6\sum _{r=1}^{m}r-\sum _{r=1}^{m}1$
 $=8{\left\{\frac{m(m+1)}{2}\right\}}^2-12\left\{\frac{m(m+1)(2m+1)}{6}\right\}+\frac{6m(m+1)}{2}-m$
 $=8{\left\{\frac{n(n+2)}{8}\right\}}^2-\frac{12}{6}\{\frac{n}{2}\left(\frac{n+2}{2}\right)+(n+1)\}+3\frac{n}{2}\left(\frac{n+2}{3}\right)-\frac{n}{2}$
 $=\frac{n^2{(n+2)}^2}{8}-\frac{n(n+1)(n+2)}{2}+3\frac{n(n+2)}{4}-\frac{n}{2}$
 $S_2=1^2+2^2+3^2+\mathrm{....}+m^2$
 $=\frac{m(m+1)(2m+1)}{6}$
 $=\frac{n(n+2)(n+1)}{24}$
 $\therefore S=S_1+12S_2$
 $=\frac{n^2{(n+2)}^2}{8}-\frac{n(n+1)(n+2)}{2}+\frac{3}{4}n(n+2)-\frac{n}{2}+\frac{n(n+1)(n+2)}{2}$$=\frac{n^2{(n+2)}^2}{8}+\frac{3}{4}n(n+2)-\frac{n}{2}$
 $=\frac{n}{8}(n^3+4n^2+10n-)$
 
Hence , $a_5=\frac{12}{37-35}=6$  is the largest positive term