Question

Sum of the following series to $n$ terms:
$2+4+7+11+16+....$

Answer 1

$S_n=2+4+7+11+16+$---n term

$S_n=2+4+7+11+$---

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$\Rightarrow t_n=2+2+3+4+5+$---nterm

$\Rightarrow t_n=2+(2+3+4+5+---nterm)$

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

$=2+\frac{n-1}{2}\{4+n-2\}$

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

$=\frac{4+n^2+2n-n-2}{2}$

$t_n=\frac{n^2+n+2}{2}$

$\therefore S_n=\sum _{n=1}^n{t}_n$

$=\frac{1}{2}\sum n^2+\frac{1}{2}\sum n+\sum _{n=1}^{n}1$

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

$=\frac{n(n+1)}{4}[\frac{2n+1}{3}+1]+n$

$=\frac{n(n+1)}{4}\frac{[2n+4]}{3}+n$

$S_n=\frac{n}{6}(n^2+3n+8)$