Question

Find the sum of $1+\frac{1+2}{2}+\frac{1+2+3}{3}+.......$ 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 $\frac{n(n-3)}{4}$

For series $1,\frac{1+2}{2},\frac{1+2+3}{3},\frac{1+2+3+4}{4},...$

$t_r=\frac{1+2+3+4+...+r}{r}=\frac{r(r+1)}{2r}=\frac{r+1}{2}=\frac{r}{2}+\frac{1}{2}$

$\therefore S_n=\sum _{r=1}^n{t}_r=\sum _{r=1}^n(\frac{r}{2}+\frac{1}{2})=\frac{1}{2}\sum _{r=1}^{n}r+\frac{1}{2}\sum _{r=1}^{n}1$

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