Question

Find the sum upto $n$ terms of the series
$3+7+13+21+31+.....nterms$.

Answer 1

Let $S_n$ be tge required sum and $T_n$ the $n^{th}$ term at series the

$S_n=3+7+13+21+31+...+T_n---\left(1\right)$

$S_n=3+7+13+21+31+...+T_{n-1}+T_n---\left(2\right)$

On subtracting $\left(2\right)$ from $\left(1\right)$

$O=3+4+6+8+10+....to(n-1)$ terms-$T_n$

$T_n=3+[4+6+8+10+...+to(n-1\left)\right]$ terms

$T_n=3+\frac{n-1}{2}\times \left[2\times 4+\left(n-1-1\right)2\right]$

$=3+\frac{n-1}{2}\times \left[8+2n-4\right]$

$=3+\frac{n-1}{2}\times \left[2n+4\right]$

$=3+\left(n-1\right)\left(n+2\right)$

$=3+n^2+2n-n-2$

$=3+n^2+n-2$

$=n^2+n+1$

$S_n=\sum n^2+\sum n+\sum 1$

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

$=\frac{n}{6}\left[\left(n+1\right)\left(2n+1\right)+3\left(n+1\right)+6\right]$

$=\frac{n}{6}\left[(2n^2+n+2n+1)+3n+3+6\right]$

Hence,

We get

$\frac{n}{6}\left[2n^2+6n+10\right]=\frac{n}{3}\left(n^2+3n+5\right)$