Find the sum of the series whose nth term is:

2n^{2} − 3n + 5

#### Solution

Let \[T_n\] be the *n*th term of the given series.

Thus, we have:

\[T_n = 2 n^2 - 3n + 5\]

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

Now,

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

\[\Rightarrow S_n = \sum^n_{k = 1} \left( 2 k^2 - 3k + 5 \right)\]

\[ \Rightarrow S_n = {2\sum}^n_{k = 1} k^2 - 3 \sum^n_{k = 1} k + \sum^n_{k = 1} 5\]

\[ \Rightarrow S_n = \frac{2n\left( n + 1 \right)\left( 2n + 1 \right)}{6} - \frac{3n\left( n + 1 \right)}{2} + 5n\]

\[ \Rightarrow S_n = \frac{2n\left( n + 1 \right)\left( 2n + 1 \right) - 9n\left( n + 1 \right) + 30n}{6}\]

\[ \Rightarrow S_n = \frac{\left( 2 n^2 + 2n \right)\left( 2n + 1 \right) - 9 n^2 - 9n + 30n}{6}\]

\[ \Rightarrow S_n = \frac{4 n^3 + 4 n^2 + 2 n^2 + 2n - 9 n^2 - 9n + 30n}{6}\]

\[ \Rightarrow S_n = \frac{4 n^3 - 3 n^2 + 23n}{6}\]

\[ \Rightarrow S_n = \frac{n\left( 4 n^2 - 3n + 23 \right)}{6}\]