1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

#### Solution

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

Thus, we have: \[T_n = n\left( n + 1 \right) = n^2 + n\]

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

Thus, we have: \[S_n = \sum^n_{k = 1} T_k\]

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

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

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

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

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

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

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