Question

Let $S_n$ denotes the sum of the cubes of the first $n$ natural numbers and $s_n$ denotes the sum of first $n$ natural numbers. Then, ${\sum }_{r=1}^n\frac{S_r}{s_r}$ equals to

• A. $\frac{n(n+1)(n+2)}{6}$
• B. $\frac{n(n+1)}{2}$
• C. $\frac{n^2+3n+2}{2}$
• D. None of these

Answer 1

Answer: (D)

$\sum _{r=1}^n\frac{S_r}{S_r}=\sum _{r=1}^n\frac{\frac{r^2(r+1{)}^2}{4}}{\frac{r(r+1)}{2}}$
$\left\{\Sigma n^3={\left[\frac{n(n+1)}{2}\right]}^2\text{ and sum of the }\text{ first }n\text{ natural numbers, }\Sigma n=\frac{n(n+1)}{2}\right\}$
$=\sum _{r=1}^n\frac{r(r+1)}{2}=\frac{1}{2}\sum _{r=1}^n\left(r^2+r\right)$
$=\frac{1}{2}\left[\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r\right]$
$=\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]$
$=\frac{1}{12}(n)(n+1)(2n+1+3)$
$=\frac{n(n+1)2(n+2)}{12}=\frac{n(n+1)(n+2)}{6}$