# 2 + 4 + 7 + 11 + 16 + ...

Question:

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

Solution:

Let $S_{n}$ be the sum of $n$ terms and $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$S_{n}=2+4+7+11+16+\ldots+T_{n-1}+T_{n}$   ....(1)

Equation (1) can be rewritten as

$S_{n}=2+4+7+11+16+\ldots+T_{n-1}+T_{n}$   ...(2)

On subtracting (2) from (1), we get:

$\Rightarrow 2+\left[\frac{(n-1)}{2}(4+(n-2) 1)\right]-T_{n}=0$

$\Rightarrow 2+\left[\frac{(n-1)}{2}(n+2)\right]-T_{n}=0$

$\Rightarrow 2+\left[\frac{n^{2}+n}{2}-1\right]-T_{n}=0$

$\Rightarrow\left[\frac{n^{2}}{2}+\frac{n}{2}+1\right]=T_{n}$

$\because S_{n}=\sum_{k=1}^{n} T_{k}$

$\therefore S_{n}=\sum_{k=1}^{n}\left(\frac{k^{2}}{2}+\frac{k}{2}+1\right)$

$=\frac{1}{2} \sum_{k=1}^{n} k^{2}+\frac{1}{2} \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$

$=\frac{n(n+1)(2 n+1)}{12}+\frac{n(n+1)}{4}+n$

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

$=\frac{n}{12}\left(2 n^{2}+6 n+16\right)$

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