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. $n^2$

Answer 1

The correct option is A

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

Let $T_n$ be the nth term of the given series Thus, we have: $T_n=\frac{1+2+3+4+5+....+n}{n}=\frac{n(n+1)}{2n}=\frac{n}{2}+\frac{1}{2}$ Now, let $S_n$ be the sum of n terms of the given series Thus, we have: $S_n={\sum }_{k=1}^n(\frac{k}{2}+\frac{1}{2})\Rightarrow S_n={\sum }_{k=1}^n\frac{k}{2}+\frac{1}{2}\Rightarrow S_n=\frac{n(n+1)}{4}+\frac{n}{2}\Rightarrow S_n=\frac{n}{2}\left(\frac{n+3}{2}\right)\Rightarrow S_n=\frac{n(n+3)}{4}$