Question

Let S_n denote the sum of the cubes of first n natural numbers and s_n denote the sum of first n natural numbers. Then, write the value of \[\sum^n_{r = 1} \frac{S_r}{s_r}\] .

Answer 1

\[\text { We know that } , S_r = 1^3 + 2^3 + 3^3 + . . . + r^3 = \left[ \frac{r\left( r + 1 \right)}{2} \right]^2 \]

\[\text { And }, s_r = 1 + 2 + 3 + . . . + r = \frac{r\left( r + 1 \right)}{2}\]

\[As, \frac{S_r}{s_r} = \frac{\left[ \frac{r\left( r + 1 \right)}{2} \right]^2}{\left[ \frac{r\left( r + 1 \right)}{2} \right]} = \frac{r\left( r + 1 \right)}{2} = \frac{1}{2}\left( r^2 + r \right)\]

Now,

\[ \sum^n_{r = 1} \frac{S_r}{s_r} = \sum^n_{r = 1} \frac{1}{2}\left( r^2 + r \right)\]

\[ = \frac{1}{2}\left( \sum^n_{r = 1} r^2 + \sum^n_{r = 1} r \right)\]

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

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

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

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

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

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