4 + 6 + 9 + 13 + 18 + ...

Question:

4 + 6 + 9 + 13 + 18 + ...

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:

$S_{n}=4+6+9+13+18+\ldots+T_{n-1}+T_{n}$   ...(1)

Equation (1) can be rewritten as:

$S_{n}=4+6+9+13+18+\ldots+T_{n-1}+T_{n}$    ...(2)

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

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

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

Now,

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

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

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

$\Rightarrow\left[\frac{n^{2}}{2}+\frac{n}{2}+3\right]=T_{n}$

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

$\therefore S_{n}=\sum_{k=1}^{n}\left(\frac{k^{2}}{2}+\frac{k}{2}+3\right)$

$=\frac{1}{2} \sum_{k=1}^{n} k^{2}+\frac{1}{2} \sum_{k=1}^{n} k+\sum_{k=1}^{n} 3$

$=\frac{n(n+1)(2 n+1)}{2 \times 6}+\frac{n(n+1)}{2 \times 2}+3 n$

$=n\left(\frac{2 n^{2}+3 n+1+3 n+3+36}{12}\right)$

$=\frac{n}{12}\left(2 n^{2}+6 n+40\right)$

$=\frac{n}{6}\left(n^{2}+3 n+20\right)$