Question

Let $S_n$ denote the sum of cubes of the first $n$ natural numbers and $s_n$ denote the sum of the first $n$ natural numbers. Then ${\sum }_{r=1}^n\frac{S_r}{S_r}$ is equal to

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

Answer 1

Answer: (A)

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

Since, $S_n$ denote the sum of cubes of the first $n$ natural numbers.
Therefore, $S_n=\frac{n^2{(n+1)}^2}{4}$
Since, $s_n$ denote the sum of the first n natural numbers.
Therefore, $s_n=\frac{n(n+1)}{2}$
Now, $T_n={\sum }_{r=1}^n\frac{S_r}{s_r}={\sum }_{r=1}^n\frac{r(r+1)}{2}$
$\Rightarrow T_n=\frac{1}{2}\{{\sum }_{r=1}^n{r}^2+{\sum }_{r=1}^{n}r\}=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}$
$\Rightarrow T_n=\frac{n(n+1)(n+2)}{6}$