If the sum of the first ten terms of the series ( 13 5 )^2 + ( 22… - Vidyadip

MCQ

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

A
$100$
B
$99$
C
$102$
✓
$101$

✓

Answer

Correct option: D.

$101$

d
$\left(1 \frac{3}{5}\right)+\left(2 \frac{2}{5}\right)^{2}+\left(3 \frac{1}{5}\right)^{2}+4^{2}+\left(4 \frac{4}{5}\right)^{2}+\ldots \ldots$ upto $10$ terms

$=\left(\frac{8}{8}\right)^{2}+\left(\frac{12}{5}\right)^{2}+\left(\frac{16}{5}\right)^{2}+\left(\frac{20}{5}\right)^{2}+\left(\frac{24}{5}\right)^{2}+\ldots \ldots \ldots \text { upto } 10 \text { terms }$

$=(8)^{2}+(12)^{2}+(16)^{2}+\ldots \ldots$ up to $10$ terms

$T_{n}=[4(n+1)]^{2}$ where $n$ varies from $1 to 10$

$=16\left(n^{2}+2 n+1\right)$

$=\left(\frac{8}{5}\right)^{2}+\left(\frac{12}{5}\right)^{2}+\left(\frac{16}{5}\right)^{2}+\left(\frac{20}{5}\right)^{2}+\left(\frac{24}{5}\right)^{2}+\ldots \ldots \ldots$ upto $10$ terms

$=\frac{16 \times 505}{25}$

$=\frac{16 \times 505}{25}=\frac{16}{5} m$

$\therefore m=\frac{505}{5}$

$=101$

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