Let $n$ th term of the series is $t_{n}$ and sum is $S$.
Then, $S=1+2+5+12+25+46+...+t_{n}$
$S=1+2+5+12+25+...+t_{n-1}+t_{n}$
On subtracting, we get
$0=1+1+3+7+13+21+...+\left(t_{n}-t_{n-1}\right)-t_{n}$
$\therefore T_{n}=1+\{1+3+7+13+21+...+$ upto $(n-1)\}$
Let $(n-1)$ th term and sum of the series $1+3+7+13+21+...$ are $t_{n-1}$ and $S'$,
respectively. Then,
$S'=1+3+7+13+21+....+t_{n-1}$
$S'=1+3+7+13+...+t_{n-2}+t_{n-1}$
On subtracting, we get
$0=1+2+4+6+8+...+\left(t_{n-1}-t_{n-2}\right)-t_{n-1}$
$\therefore t_{n-1} =1+2\{1+2+3+4+... \text { upto }(n-2)\}$
$=1+2 \cdot \frac{1}{2}(n-2)(n-1)=n^{2}-3 n+3$
$\Rightarrow t_{n} =(n+1)^{2}-3(n+1)+3$
$=n^{2}-n+1$
$\therefore t_{n} =1+\{1+3+7+13+... \text { upto }(n-1)\}$
$=1+\displaystyle \sum_{n=1}^{n-1}\left(n^{2}-n+1\right)$
$=1+\displaystyle \sum_{n=1}^{n-1} n^{2}-\displaystyle \sum_{n=1}^{n-1} n+\displaystyle \sum_{n=1}^{n-1} 1$
$=1+\frac{1}{6} n(n-1)(2 n-1)$
$-\frac{1}{2} n(n-1)+(n-1)$
$=\frac{1}{3} n(n-1)(n-2)+n$
Hence, $t_{n}=\frac{1}{3} n(n-1)(n-2)+n$