# 1 + 4 + 13 + 40 + 121 + ...

Question:

1 + 4 + 13 + 40 + 121 + ...

Solution:

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

Thus, we have:

$S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n}$   ....(1)

Equation (1) can be rewritten as:

$S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n}$    ...(2)

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

The sequence of difference between successive terms is 3, 9, 27, 81,...

We observe that it is a GP with common ratio 3 and first term 3.

Thus, we have:

$1+\left[\frac{3\left(3^{n-1}-1\right)}{3-1}\right]-T_{n}=0$

$\Rightarrow 1+\left[\frac{\left(3^{n}-3\right)}{2}\right]-T_{n}=0$

$\Rightarrow\left(\frac{3^{n}}{2}-\frac{1}{2}\right)-T_{n}=0$

$\Rightarrow\left(\frac{3^{n}}{2}-\frac{1}{2}\right)=T_{n}$

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

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

$\Rightarrow S_{n}=\frac{1}{2} \sum_{k=1}^{n} 3^{k}-\frac{1}{2} \sum_{k=1}^{n} 1$

$\Rightarrow S_{n}=\frac{1}{2}\left(3+3^{2}+3^{3}+3^{4}+3^{5}+\ldots+3^{n}\right)-\frac{n}{2}$

$\Rightarrow S_{n}=\frac{1}{2}\left[\frac{3\left(3^{n}-1\right)}{2}\right]-\frac{n}{2}$

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

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