Mathematics - Sequence & Series Question with Solution | TestHub

We have:

$$\begin{array}{rl}S& =1^3+3\cdot 2^2+3^3+3\cdot 4^2+5^3+3\cdot 6^2+\dots \\ S& =(1^3+3^3+5^3+\dots )+3\cdot (2^2+4^2+6^2+\dots )\\ S& =(1^3+3^3+5^3+\dots )+12(1^2+2^2+3^2+\dots )\end{array}$$

Where $S_1=1^3+3^3+5^3+\dots$ and $S_2=1^2+2^2+3^2+\dots$. Now, a case arises when $n$ is, say, even, for example, $n=2m$, where $m\in N$. In this case, $S_1$ and $S_2$ both contain $m$ terms.

$$\begin{array}{rl}\therefore S_1& =1^3+3^3+5^3+\dots +(2m-1{)}^3\\ & =\sum _{r=1}^m(2r-1{)}^3\\ & =\sum _{r=1}^m(8r^3-12r^2+6r-1)\\ & =8\sum _{r=1}^m{r}^3-12\sum _{r=1}^m{r}^2+6\sum _{r=1}^{m}r-\sum _{r=1}^{m}1\\ & =8{\left\{\frac{m(m+1)}{2}\right\}}^2-12\left\{\frac{m(m+1)(2m+1)}{6}\right\}+\frac{6m(m+1)}{2}-m\end{array}$$

$$=8{\left\{\frac{n(n+2)}{8}\right\}}^2-\frac{12}{6}\left\{\frac{n}{2}\left(\frac{n+2}{2}\right)+(n+1)\right\}+3\frac{n}{2}\left(\frac{n+2}{3}\right)-\frac{n}{2}$$

$$=\frac{n^2(n+2{)}^2}{8}-\frac{n(n+1)(n+2)}{2}+\frac{3n(n+2)}{4}-\frac{n}{2}$$

$S_2=1^2+2^2+3^2+\dots +m^2$

$$=\frac{m(m+1)(2m+1)}{6}$$

$$=\frac{n(n+2)(n+1)}{24}$$

$\therefore S=S_1+12S_2$

$$=\frac{n^2(n+2{)}^2}{8}-\frac{n(n+1)(n+2)}{2}+\frac{3}{4}n(n+2)-\frac{n}{2}+\frac{n(n+1)(n+2)}{2}$$

$$=\frac{n^2(n+2{)}^2}{8}+\frac{3}{4}n(n+2)-\frac{n}{2}$$

$$=\frac{n}{8}\left(n^3+4n^2+10n-\right)$$