Question

If the sum of the first ten terms of the series

${\left(1\frac{3}{5}\right)}^2+{\left(2\frac{2}{5}\right)}^2+{\left(3\frac{1}{5}\right)}^2+4^2+{\left(4\frac{4}{5}\right)}^2+.....,$

Is $\frac{16}{5}$ m, then m is equal to:

• A. 101
• B. 100
• C. 99
• D. 102

Answer 1

The correct option is A

101

${\left(1\frac{3}{5}\right)}^2+{\left(2\frac{2}{5}\right)}^2+{\left(3\frac{1}{5}\right)}^2+4^2+......=\frac{16}{5}m$

$\Rightarrow {\left(\frac{8}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2+{\left(\frac{16}{5}\right)}^2+{\left(\frac{20}{5}\right)}^2+......10terms=\frac{16}{5}m$

$\Rightarrow {\left(\frac{4}{5}\right)}^2[2^2+3^2+4^2+5^2......10terms]=\frac{16}{5}m$

$\Rightarrow {\left(\frac{4}{5}\right)}^2[2^2+3^2+4^2......+{11}^2]=\frac{16}{5}m$

$\Rightarrow {\left(\frac{4}{5}\right)}^2[1^2+2^2+3^2......+{11}^2-1^2]=\frac{16}{5}m$

$\Rightarrow {\left(\frac{4}{5}\right)}^2[\frac{\mathrm{11.12.23}}{6}-1]=\frac{16}{5}m\left(given\right)$

$\Rightarrow \frac{16}{25}[22.23-1]=\frac{16}{5}m$

$\Rightarrow \frac{1}{5}\left(505\right)=m$

$\Rightarrow M=101$