Question

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

Answer 1

Let $T_n$ be the nth term and $S_n$ be the sum to n terms of the given series.

Thus, we have:

$S_n=1+4+13+40+121+...+T_{n-1}+T_n$ ...(1)

Equation (1) can be rewritten as:

$S_n=1+4+13+40+121+...+T_{n-1}+T_n$ ...(2)

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

$$\begin{gathered} S_n=1+4+13+40+121+...+T_{n-1}+T_n \\ \underline{\underline{S_n=1+4+13+40+121+...+T_{n-1}+T_n}} \\ 0=1+\left[3+9+27+81+...+\left(T_n-T_{n-1}\right)\right]-T_n \end{gathered}$$

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:

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

$$\begin{gathered} \because S_n=\sum _{k=1}^n{T}_k \\ \therefore S_n=\sum _{k=1}^n\left(\frac{3^k}{2}-\frac{1}{2}\right) \\ \Rightarrow S_n=\frac{1}{2}\sum _{k=1}^n{3}^k-\frac{1}{2}\sum _{k=1}^{n}1 \\ \Rightarrow S_n=\frac{1}{2}\left(3+3^2+3^3+3^4+3^5+...+3^n\right)-\frac{n}{2} \\ \Rightarrow S_n=\frac{1}{2}\left[\frac{3\left(3^n-1\right)}{2}\right]-\frac{n}{2} \\ \Rightarrow S_n=\left(\frac{3^{n+1}-3}{4}\right)-\frac{n}{2} \\ \Rightarrow S_n=\left(\frac{3^{n+1}-3-2n}{4}\right) \end{gathered}$$