Mathematics - Sequence & Series Question with Solution | TestHub

$S_{\mathrm{20}}=\frac{3}{{\mathrm{1}}^2}+\frac{5}{{\mathrm{1}}^2+2^2}+\frac{7}{1^2+2^2+3^2}+....+\mathrm{upto} 20terms$

$=\underset{r=1}{\overset{\mathrm{20}}{\Sigma }}\frac{\left(2r+1\right)}{{\mathrm{1}}^2+2^2+3^2+.... r^2}$

$=\underset{r=1}{\overset{\mathrm{20}}{\Sigma }}\frac{2r+1}{\frac{r}{6}\left(r+1\right)\left(2r+1\right)}$

$=6\underset{r=1}{\overset{\mathrm{20}}{\Sigma }}\frac{1}{r\left(r+1\right)}$

$=6\underset{r=1}{\overset{\mathrm{20}}{\Sigma }}\left(\frac{1}{r}-\frac{1}{r+1}\right)$

$=6\left\{\left[\frac{1}{1}-\frac{1}{2}\right]+\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{4}\right]+....+\left[\frac{1}{\mathrm{20}}-\frac{1}{\mathrm{21}}\right]\right\}$

$=6\left(1-\frac{1}{\mathrm{21}}\right)$

$=\frac{6\times \mathrm{20}}{\mathrm{21}}$

$=\frac{120}{\mathrm{21}}$

$=\frac{k}{\mathrm{21}}$

$k=120$