Question

If the sum of the first ten term 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 $\frac{16m}{5}$, then m is equal to

• A. $607$
• B. $200$
• C. $99$
• D. $102$

Answer 1

The correct option is A $607$

${\left(1\frac{3}{5}\right)}^2+{\left(2\frac{2}{5}\right)}^2+{\left(3\frac{1}{5}\right)}^2+{\left(4\right)}^2+....+$ to $10$ terms is $\frac{16m}{5}$

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

$\Rightarrow \frac{8^2+{12}^2+{16}^2+...}{5^2}$ to $10$ terms is $\frac{16m}{5}$

$\Rightarrow \frac{4^2(2^2+3^2+4^2+.....)}{5^2}$ to $10$ terms is $\frac{16m}{5}$

$\Rightarrow \frac{4^2}{5^2}(1^2+2^2+3^2+4^2+....-1^2)$ to $10$ terms is $\frac{16m}{5}$

$\Rightarrow \frac{16}{25}{\sum }_{r=1}^{10}{r}^2-\frac{16}{25}\times 1$

$=\frac{16}{25}\frac{10(10+1)(2\times 10+1)}{6}-\frac{16}{25}$ where ${\sum }_{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6}$

$=\frac{16\times 3036}{25}-\frac{16}{25}=\frac{16m}{5}$(given)

$\Rightarrow \frac{16}{25}(3036-1)=\frac{16m}{5}$

$\Rightarrow \frac{16}{25}\times 3035=\frac{16m}{5}$

$\Rightarrow m=607$ on simplification

$\therefore m=607$