2 + 5 + 10 + 17 + 26 + ...

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

Thus, we have:

$S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$    ...(1)

Equation (1) can be rewritten as:

$S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$   ...(2)

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

$S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$

$S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$

$0=2+\left[3+5+7+9+\ldots+\left(T_{n}-T_{n-1}\right)\right]-T_{n}$

The sequence of difference of successive terms is 3, 5, 7, 9,...

We observe that it is an AP with common difference 2 and first term 3.

Thus, we have:

$2+\left[\frac{(n-1)}{2}\{6+(n-2) 2\}\right]-T_{n}=0$

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

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

Now,

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

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

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

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

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

$\Rightarrow S_{n}=\frac{n\left(2 n^{2}+3 n+7\right)}{6}$