1 + 3 + 6 + 10 + 15 + ...

Solution:

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

Thus, we have:

$S_{n}=1+3+6+10+15+\ldots+T_{n-1}+T_{n}$   ...(1)

Equation (1) can be rewritten as:

$S_{n}=1+3+6+10+15+\ldots+T_{n-1}+T_{n}$   ....(2)

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

The sequence of difference of successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Thus, we have:

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

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

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

Now,

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

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

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

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

$\Rightarrow S_{n}=\frac{n(n+1)}{4}\left(\frac{2 n+1}{3}+1\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{4}\left(\frac{2 n+4}{3}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{n+2}{3}\right)$

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