Question

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

(a) $\frac{n(n+3)}{4}$

(b) $\frac{n(n+2)}{4}$

(c) $\frac{n(n+1)(n+2)}{6}$

(d) n²

Answer 1

$\left(\mathrm{a}\right)\frac{n(n+3)}{4}$

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:

$$\begin{gathered} S_n=\sum _{k=1}^n\left(\frac{k}{2}+\frac{1}{2}\right) \\ \Rightarrow S_n=\sum _{k=1}^n\frac{k}{2}+\frac{n}{2} \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{4}+\frac{n}{2} \\ \Rightarrow S_n=\frac{n}{2}\left(\frac{n+1}{2}+1\right) \\ \Rightarrow S_n=\frac{n}{2}\left(\frac{n+3}{2}\right) \\ \Rightarrow S_n=\frac{n\left(n+3\right)}{4} \end{gathered}$$