Question

If \[1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .\] to n terms is S, then S is equal to

Options

• \[\frac{n (n + 3)}{4}\]

• \[\frac{n (n + 2)}{4}\]

• \[\frac{n (n + 1) (n + 2)}{6}\]

•  n²

Answer 1

\[\frac{n (n + 3)}{4}\]

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

\[T_n = \frac{1 + 2 + 3 + 4 + 5 + . . . + n}{n} = \frac{n\left( n + 1 \right)}{2n} = \frac{n}{2} + \frac{1}{2}\]

Now, let

\[S_n\]  be the sum of n terms of the given series.
Thus, we have:

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

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

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

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

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

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