Question

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...

Answer 1

Let \[T_n\] be the nth term of the given series.
Thus, we have: \[T_n = 1 + 2 + 3 + 4 + 5 + . . . + n = \frac{n\left( n + 1 \right)}{2} = \frac{n^2 + n}{2}\]

Now, let

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

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

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

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

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

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

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

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 4 \right)}{12}\]

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