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. 100
• B. 99
• C. 102
• D. 101

Answer 1

The correct option is D 101

${\left(\frac{8}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2+{\left(\frac{16}{5}\right)}^2+{\left(\frac{20}{5}\right)}^2...+{\left(\frac{44}{5}\right)}^{2}S=\frac{16}{25}(2^2+3^2+4^2+...+{11}^2)=\frac{16}{25}(\frac{11(11+1)(22+1)}{6}-1)=\frac{16}{25}\times 505=\frac{16}{5}\times 101\Rightarrow \frac{16}{5}m=\frac{16}{5}\times 101\Rightarrow m=101.$