Question

Find the sum of first $n$ terms of the series.
$1^3+3\times 2^2+3^3+3\times 4^2+5^3+3\times 6^2+....$ If $n$ is even.

• A. $\frac{n}{8}(n^3+4n^2+10n+6)$
• B. $\frac{n}{8}(n^2+4n^2+10n+6)$
• C. $\frac{n}{8}(n^3+4n^2+10n+8)$
• D. $\frac{n}{8}(n^2+4n^2+10n+8)$

Answer 1

The correct option is C $\frac{n}{8}(n^3+4n^2+10n+8)$

If $n$ is even,
Let $n=2m$
$\therefore S=1^3+{3.2}^2+3^3+{3.4}^2+5^3+{3.6}^2+....+(2m-1{)}^3+3(2m{)}^2$
$=\{1^3+3^3+5^3+...+(2m-1{)}^3\}+3\{2^2+4^2+6^2+....+(2m{)}^2\}$
$={\sum }_{r=1}^m(2r-1{)}^3+3.4{\sum }_{r=1}^m{r}^2$
$={\sum }_{r=1}^m\{8r^3-12r^2+6r-1\}+12{\sum }_{r=1}^m{r}^2$
$=8{\sum }_{r=1}^m{r}^3-12{\sum }_{r=1}^m{r}^2+6{\sum }_{r=1}^{m}r-{\sum }_{r=1}^{m}1+12{\sum }_{r=1}^m{r}^2$
$=8{\sum }_{r=1}^m{r}^3+6{\sum }_{r=1}^{m}r-{\sum }_{r=1}^{m}1$
$=8.\frac{m^2(m+1{)}^2}{4}+6\frac{m(m+1)}{2}-m$
$=2m^2(m+1{)}^2+3m(m+1)-m$
$=m[2m^3+4m^2+5m+2]$
$=\frac{n}{2}[2{\left(\frac{n}{2}\right)}^3+4{\left(\frac{n}{2}\right)}^2+5\left(\frac{n}{2}\right)+2]$ $(\because m=\frac{n}{2})$
$=\frac{n}{8}(n^3+4n^2+10n+8)$