# Solve the following

Question:

Let $S_{n}$ denote the sum of the cubes of the first $n$ natural numbers and $S_{n}$ denote the sum of the first natural numbers. Then $\sum_{r=1}^{n} \frac{S_{r}}{S_{r}}$ equals

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

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

(c) $\frac{n^{2}+3 n+2}{2}$

(d) None of these

Solution:

Let $S_{n}$ denote the sum of the cubes of the first $n$ natural numbers and $S_{n}$ denotes the sum of the first $n$ natural number 1 .

Then $\sum_{r=1}^{n} \frac{S_{r}}{S_{r}}=?$

Since $S_{n}=\left[\frac{n(n+1)}{2}\right]^{2} ;$ formula for sum of cubes.

$S_{n}=\frac{n(n+1)}{2} \quad ;$ formula form sum of first $n$ natural numbers.

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

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

$=\frac{1}{2}\left[\sum_{r=1}^{n} r^{2}+\sum_{r=1}^{n} r\right]$

$=\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]$

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

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

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

$=\frac{n(n+1)[2(n+2)]}{12}$

Hence, $\sum_{r=1}^{n} \frac{S_{r}}{S_{n}}=\frac{n(n+1)(n+2)}{6}$

Hence, the correct answer is option A.