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

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Question

(C) The given series is: $\left(\frac{11}{7}\right)^{2} + \left(\frac{12}{7}\right)^{2} + \left(\frac{13}{7}\right)^{2} + \ldots$ The $n^{th}$ term is

Vedclass · Accepted Answer

$38$

Vedclass · Answer

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

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

Similar Questions

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