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. 102
• B. 101
• C. 100
• D. 99

Answer 1

Answer: (B)

$\left(1 \frac{3}{5}\right)^{2} + \left(2 \frac{2}{5}\right)^{2} + \left(3 \frac{1}{5}\right)^{2} + 4^{2} + \left(4 \frac{4}{5}\right)^{2} + .....$ upto $10$ terms
$ = \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{24}{5}\right)^{2} + ... $ upto $10$ terms.
$\left(8\right)^{2} + \left(12\right)^{2} + \left(16\right)^{2} + ...$ up to $10$ terms
$ T_{n} \left[4\left(n+1\right)\right]^{2} $ where n varies from $1$ to $10$.
$= 16\left(n^{2} + 2n + 1\right) $
$\displaystyle\sum T_{n} = \sum^{10}_{n=1} 16 \left(n^{2} + 2n + 1\right)$
$ = 16 \left[385 + 55\left(2\right) + 10\right] $
$= 16\left(505\right) $
or
$\displaystyle\sum^{10}_{n=1} n^{2} = \frac{n\left(n+1\right)\left(2n+1\right)}{6} = \frac{10 \times11 \times21}{6} = 385 $
$\displaystyle\sum^{10}_{n=1} n = \frac{n\left(n+1\right)}{2} = \frac{10 \times11}{2} = 55$
$\displaystyle \sum^{10}_{n=1} 1 = n = 10 $
$\therefore \left(\frac{8}{5}\right)^{2} + \left(\frac{12}{5}\right)^{2} + \left(\frac{16}{5}\right)^{2} + ..... $ upto $10$ terms $= \frac{16 \times505}{25 }$
It is given that $\frac{16 \times505}{25} = \frac{16}{5}\, m$
$ \therefore m = \frac{505}{5} = 101$