Question

Find the $nth$ term and sum of $n$ terms of the series $2+4+7+11+16+....$

Answer 1

Given ,

$2+4+7+11+16+....$

$\begin{array}{c}{T}_n=a{n}^2+bx+c\\ T_1=a+b+c=\mathrm{2...........}\left(i\right)3a+b=2\\ T_2=4a+2b+c=\mathrm{4...........}\left(ii\right)5a+b=3\\ T_3=9a+3b+c=\mathrm{7...........}\left(iii\right)2a=\\ a=\frac{1}{2},b=2-\frac{3}{2}=\frac{1}{2},c=1\\ T_n=\frac{n^2}{2}+\frac{n}{2}+1\left(n^{th}term\right)\end{array}$

Sum to $n$ terms $=S_n=\Sigma \frac{r^2}{2}+\Sigma \frac{r}{2}+\Sigma 1$

$\begin{array}{c}=\frac{1}{2}\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{1}{2}\frac{n\left(n+1\right)}{2}+n\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{12}+\frac{n\left(n+1\right)}{4}+n\\ =\frac{\left(n^2+n\right)\left(2n+1\right)+3n\left(n+1\right)+12n}{12}\\ =\frac{2n^3+6x^2+16n}{12}\\ =\frac{n^3+3n^2+8n}{6}\end{array}$