If the sum of the first ten terms of the seri | vedclass

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(D) The given series is ${\left( \frac{8}{5} \right)^2} + {\left( \frac{12}{5} \right)^2} + {\left( \frac{16}{5} \right)^2} + {\left( \frac{20}{5} \ri

Vedclass · Accepted Answer

$101$

Vedclass · Answer

(D) The given series is ${\left( \frac{8}{5} ight)^2} + {\left( \frac{12}{5} ight)^2} + {\left( \frac{16}{5} ight)^2} + {\left( \frac{20}{5} ight)^2} + \dots$ up to $10$ terms.This can be written as $\frac{1}{25} [8^2 + 12^2 + 16^2 + 20^2 + \dots + (4(n+1))^2]$ for $n=1$ to $10$.The $n$-th term is $T_n = {\left( \frac{4(n+1)}{5} ight)^2} = \frac{16}{25} (n^2 + 2n + 1)$.The sum of $10$ terms is $S_{10} = \frac{16}{25} \sum_{n=1}^{10} (n^2 + 2n + 1)$.Using the formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum n = \frac{n(n+1)}{2}$,and $\sum 1 = n$:$S_{10} = \frac{16}{25} [\frac{10(11)(21)}{6} + 2 \cdot \frac{10(11)}{2} + 10] = \frac{16}{25} [385 + 110 + 10] = \frac{16}{25} [505]$.Given $S_{10} = \frac{16}{5} m$,we have $\frac{16}{25} 	imes 505 = \frac{16}{5} m$.Dividing both sides by $\frac{16}{5}$,we get $m = \frac{505}{5} = 101$.

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

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