Question

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

• A. 102
• B. 101
• C. 100
• D. 99

Answer 1

The correct option is B 101

Let $S_{10}$ be the sum of first ten terms of the series. Then, we have
$S_{10}={\left(1\frac{3}{5}\right)}^2+{\left(2\frac{2}{5}\right)}^2+{\left(3\frac{1}{5}\right)}^2+4^2+{\left(4\frac{4}{5}\right)}^2+\dots \dots to10terms$
$={\left(\frac{8}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2+{\left(\frac{16}{5}\right)}^2+4^2+{\left(\frac{24}{5}\right)}^2+\dots \dots to10terms$
$=\frac{1}{5^2}(8^2+{12}^2+{16}^2+{20}^2+{24}^2+\dots )$
$=\frac{4^2}{5^2}(2^2+3^3+4^2+5^2+\dots \dots )$
$=\frac{4^2}{5^2}(2^2+3^3+4^2+5^2+\dots \dots +{11}^2)$
$=\frac{16}{25}\left(\right(1^2+2^2+\dots \dots +{11}^2)-1^2)$
$=\frac{16}{25}(\frac{11.(11+1)(2.11+1)}{6}-1)$
$=\frac{16}{25}(506-1)=\frac{16}{25}\times 505\Rightarrow \frac{16}{5}m=\frac{16}{5}\times 101$
$\Rightarrow m=101$