Question

# Let Sn denote the sum of the cubes of first n natural numbers and sn denote the sum of first n natural numbers. Then, write the value of $\sum _{r=1}^{n}\frac{{S}_{r}}{{s}_{r}}$.

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Solution

## $\mathrm{We}\mathrm{know}\mathrm{that},{S}_{r}={1}^{3}+{2}^{3}+{3}^{3}+...+{r}^{3}={\left[\frac{r\left(r+1\right)}{2}\right]}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{And},{s}_{r}=1+2+3+...+r=\frac{r\left(r+1\right)}{2}\phantom{\rule{0ex}{0ex}}\mathrm{As},\frac{{S}_{r}}{{s}_{r}}=\frac{{\left[\frac{r\left(r+1\right)}{2}\right]}^{2}}{\left[\frac{r\left(r+1\right)}{2}\right]}=\frac{r\left(r+1\right)}{2}=\frac{1}{2}\left({r}^{2}+r\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\sum _{r=1}^{n}\frac{{S}_{r}}{{s}_{r}}=\sum _{r=1}^{n}\frac{1}{2}\left({r}^{2}+r\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\sum _{r=1}^{n}{r}^{2}+\sum _{r=1}^{n}r\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{n\left(n+1\right)}{2}×\left[\frac{\left(2n+1\right)}{3}+1\right]\phantom{\rule{0ex}{0ex}}=\frac{n\left(n+1\right)}{4}\left[\frac{2n+1+3}{3}\right]\phantom{\rule{0ex}{0ex}}=\frac{n\left(n+1\right)}{4}\left[\frac{2n+4}{3}\right]\phantom{\rule{0ex}{0ex}}=\frac{n\left(n+1\right)}{4}×\frac{2\left(n+2\right)}{3}\phantom{\rule{0ex}{0ex}}=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

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