Question

3 × 1² + 5 ×2² + 7 × 3² + ...

Answer 1

Let $T_n$ be the nth term of the given series.

Thus, we have:

$T_n=\left(2n+1\right)n^2=2n^3+n^2$

Now, let $S_n$ be the sum of n terms of the given series.

Thus, we have:
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n\left(2k^3+k^2\right) \\ \Rightarrow S_n=\underset{k=1}{\overset{n}{2\sum }}{k}^3+\sum _{k=1}^n{k}^2 \\ \Rightarrow S_n=\left[\frac{2n^2{\left(n+1\right)}^2}{4}+\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right] \\ \Rightarrow S_n=\left[\frac{n^2{\left(n+1\right)}^2}{2}+\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left[n\left(n+1\right)+\frac{2n+1}{3}\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{3n^2+3n+2n+1}{3}\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{3n^2+5n+1}{3}\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{6}\left(3n^2+5n+1\right) \end{gathered}$$