Question

Find the sum of the series whose nth term is:

 2n³ + 3n² − 1

Answer 1

Let \[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = 2 n^3 + 3 n^2 - 1\]

Let \[S_n\] be the sum of n terms of the given series.
Now,

\[S_n = \sum^n_{k = 1} T_k\]

\[\Rightarrow S_n = \sum^n_{k = 1} \left( 2 k^3 + 3 k^2 - 1 \right)\]

\[ \Rightarrow S_n = {2\sum}^n_{k = 1} k^3 + 3 \sum^n_{k = 1} k^2 - \sum^n_{k = 1} 1\]

\[ \Rightarrow S_n = \frac{2 n^2 \left( n + 1 \right)^2}{4} + \frac{3n\left( n + 1 \right)\left( 2n + 1 \right)}{6} - n\]

\[ \Rightarrow S_n = \frac{n^2 \left( n + 1 \right)^2}{2} + \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{2} - n\]

\[ \Rightarrow S_n = \frac{n^2 \left( n + 1 \right)^2 + n\left( n + 1 \right)\left( 2n + 1 \right) - 2n}{2}\]

\[ \Rightarrow S_n = \frac{n^2 \left( n^2 + 1 + 2n \right) + \left( n^2 + n \right)\left( 2n + 1 \right) - 2n}{2}\]

\[ \Rightarrow S_n = \frac{\left( n^4 + n^2 + 2 n^3 \right) + \left( 2 n^3 + n^2 + 2 n^2 + n \right) - 2n}{2}\]

\[ \Rightarrow S_n = \frac{n^4 + 4 n^2 + 4 n^3 - n}{2}\]

\[ \Rightarrow S_n = \frac{n\left( n^3 + 4n + 4 n^2 - 1 \right)}{2}\]