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,
$\because S_{n}=\sum_{k=1}^{n} T_{k}$
$\therefore S_{n}=\sum_{k=1}^{n}\left(k^{2}+1\right)$
$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 1$
$\Rightarrow S_{n}=\frac{n(n+1)(2 n+1)}{6}+n$
$\Rightarrow S_{n}=\frac{n(n+1)(2 n+1)+6 n}{6}$
$\Rightarrow S_{n}=\frac{n\left(2 n^{2}+3 n+7\right)}{6}$
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