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 $101$

This can be written as

$(\frac{8}{5}{)}^2+(\frac{12}{5}{)}^2+(\frac{16}{5}{)}^2+(\frac{20}{5}{)}^2+......$

We can see that the numbers are in AP

$a_{10}={10}^{th}$ term $=\frac{8}{5}+\frac{4}{5}\times (10-1)=\frac{44}{5}$

Hence, Sum of first ten terms is

$(\frac{8}{5}{)}^2+(\frac{12}{5}{)}^2+(\frac{16}{5}{)}^2+(\frac{20}{5}{)}^2+......(\frac{44}{5}{)}^2$

Common term is $a_k=\frac{8}{5}+(k-1)\frac{4}{5}=\frac{4}{5}(k+1)$

${\sum }_{k=1}^n\left[\frac{4}{5}\right(k+1){]}^2=\frac{16}{25}[{\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}2k+{\sum }_{k=1}^{n}1]=\frac{16}{25}[\frac{\left(n\right)(n+1)(2n+1)}{6}+2\times \frac{\left(n\right)(n+1)}{2}+n]$

and so,

${\sum }_{k=1}^{10}\left[\frac{4}{5}\right(k+1){]}^2=\frac{16}{25}[\frac{\left(10\right)(10+1)(2\times 10+1)}{6}+2\times \frac{\left(10\right)(10+1)}{2}+10]=\frac{16\times 505}{25}=\frac{16}{5}\times 101$

Therefore, $m=101$