Question

If the sum of the first ten terms of the series
${\left(1\frac{3}{5}\right)}^2+{\left(2\frac{2}{5}\right)}^2+{\left(3\frac{1}{5}\right)}^2+4^2+{\left(4\frac{4}{5}\right)}^2+.......,$
$\frac{16}{5}m$, then m is equal to

Answer 1

$\begin{array}{c}S={\left(1\frac{3}{5}\right)}^2+{\left(2\frac{2}{5}\right)}^2+{\left(3\frac{1}{5}\right)}^2+4^2+{\left(4\frac{4}{5}\right)}^2+............\\ S={\left(\frac{8}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2+{\left(\frac{16}{5}\right)}^2+{\left(\frac{20}{5}\right)}^2+{\left(\frac{24}{5}\right)}^2+............\\ S=\frac{{\sum }_{r=2}^{11}{\left(4r\right)}^2}{{\left(5\right)}^2}\\ S=\frac{16}{25}{\sum }_{r=2}^{11}r\\ =\frac{16}{25}\left[{\sum }_{r=1}^{11}{r}^2-1\right]\\ =\frac{16}{25}\left[\frac{11\left(12\right)\left(23\right)}{6}-1\right]\\ =\frac{16}{25}\left[22\times 23-1\right]\\ =\frac{16}{25}\left[505\right]\\ =\frac{16}{5}\left(101\right)\\ \therefore m=101\end{array}$