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

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(D) The series is $S = \left(\frac{8}{5}\right)^2 + \left(\frac{12}{5}\right)^2 + \left(\frac{16}{5}\right)^2 + \left(\frac{20}{5}\right)^2 + \dots$ u

Vedclass · Accepted Answer

$101$

Vedclass · Answer

(D) The series is $S = \left(\frac{8}{5}ight)^2 + \left(\frac{12}{5}ight)^2 + \left(\frac{16}{5}ight)^2 + \left(\frac{20}{5}ight)^2 + \dots$ up to $10$ terms.This can be written as $S = \frac{1}{25} \left( 8^2 + 12^2 + 16^2 + 20^2 + \dots + (4(n+1))^2 ight)$ for $n=1$ to $10$.$S = \frac{16}{25} \sum_{n=1}^{10} (n+1)^2$.Let $k = n+1$,then the sum is $\frac{16}{25} \sum_{k=2}^{11} k^2$.Using the formula $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$,we have $\sum_{k=1}^{11} k^2 = \frac{11(12)(23)}{6} = 11 	imes 2 	imes 23 = 506$.Since the sum starts from $k=2$,we subtract $1^2 = 1$: $506 - 1 = 505$.So,$S = \frac{16}{25} 	imes 505 = \frac{16 	imes 101}{5} = \frac{16}{5} 	imes 101$.Given $S = \frac{16}{5}m$,we get $m = 101$.

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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