If the sum of the first 11 terms of the serie | vedclass

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Question

(C) The given series is $\left(\frac{11}{7}\right)^2 + \left(\frac{12}{7}\right)^2 + \left(\frac{13}{7}\right)^2 + \dots + \left(\frac{21}{7}\right)^2

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(C) The given series is $\left(\frac{11}{7}ight)^2 + \left(\frac{12}{7}ight)^2 + \left(\frac{13}{7}ight)^2 + \dots + \left(\frac{21}{7}ight)^2$.The sum of the first $11$ terms is $S = \sum_{n=11}^{21} \left(\frac{n}{7}ight)^2 = \frac{1}{49} \sum_{n=11}^{21} n^2$.Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we have:$S = \frac{1}{49} \left[ \sum_{n=1}^{21} n^2 - \sum_{n=1}^{10} n^2 ight]$$S = \frac{1}{49} \left[ \frac{21 	imes 22 	imes 43}{6} - \frac{10 	imes 11 	imes 21}{6} ight]$$S = \frac{1}{49} 	imes \frac{21}{6} 	imes [ (22 	imes 43) - (10 	imes 11) ]$$S = \frac{1}{49} 	imes \frac{7}{2} 	imes [ 946 - 110 ]$$S = \frac{1}{14} 	imes 836 = \frac{418}{7} = \frac{11}{7} 	imes 38$.Comparing this with $\frac{11}{7}\lambda$,we get $\lambda = 38$.

If the sum of the first $11$ terms of the series ${\left( {1\frac{4}{7}} \right)^2} + {\left( {1\frac{5}{7}} \right)^2} + {\left( {1\frac{6}{7}} \right)^2} + {2^2} + {\left( {2\frac{1}{7}} \right)^2} + \dots$ is $\frac{11}{7}\lambda$,then $\lambda$ is equal to:

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