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

$S_n=1+\frac{1+2}{2}+\frac{1+2+3}{3}+.........$to $n$ terms

We have to find $r^{th}$ term of two sequence

$T_r=\frac{r\left(r+1\right)}{2r}=\frac{r+1}{2}$

$S_n=\sum _{n-1}^n{T}_r=\sum _{n=1}^n\left(\frac{r+1}{2}\right)=\frac{1}{2}\left(\sum r+\sum 1\right)$

$\frac{1}{2}\left[\frac{n\left(r+1\right)}{2}+n\right]$

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

Hence option (A) is correct$.$