Question

2 + 4 + 7 + 11 + 16 + ...

Answer 1

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

Thus, we have:

$S_n=2+4+7+11+16+...+T_{n-1}+T_n$ ...(1)

Equation (1) can be rewritten as:

$S_n=2+4+7+11+16+...+T_{n-1}+T_n$ ...(2)

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

$$\begin{gathered} S_n=2+4+7+11+16+...+T_{n-1}+T_n \\ {S}_n=2+4+7+11+16+...+T_{n-1}+T_n \\ \underline{--------} \\ 0=2+\left[2+3+4+5+6+...+\left(T_n-T_{n-1}\right)\right]-T_n \end{gathered}$$

$$\begin{gathered} \Rightarrow 2+\left[\frac{\left(n-1\right)}{2}\left(4+\left(n-2\right)1\right)\right]-T_n=0 \\ \Rightarrow 2+\left[\frac{\left(n-1\right)}{2}\left(n+2\right)\right]-T_n=0 \\ \Rightarrow 2+\left[\frac{n^2+n}{2}-1\right]-T_n=0 \\ \Rightarrow \left[\frac{n^2}{2}+\frac{n}{2}+1\right]=T_n \end{gathered}$$

$$\begin{gathered} \because S_n=\sum _{k=1}^n{T}_k \\ \therefore S_n=\sum _{k=1}^n\left(\frac{k^2}{2}+\frac{k}{2}+1\right) \\ =\frac{1}{2}\sum _{k=1}^n{k}^2+\frac{1}{2}\sum _{k=1}^{n}k+\sum _{k=1}^{n}1 \\ =\frac{n\left(n+1\right)\left(2n+1\right)}{12}+\frac{n\left(n+1\right)}{4}+n \\ =n\left(\frac{2n^2+3n+1+3n+3+12}{12}\right) \\ =\frac{n}{12}\left(2n^2+6n+16\right) \\ =\frac{n}{6}\left(n^2+3n+8\right) \end{gathered}$$