The sum frac{3}{1^2} + frac{5}{1^2 + 2^2} + f | vedclass

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(C) The given series is $\sum_{n=1}^{11} T_n$,where the $n^{th}$ term $T_n$ is given by:$T_n = \frac{2n+1}{1^2 + 2^2 + \dots + n^2}$Using the formula

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$\frac{11}{2}$

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(C) The given series is $\sum_{n=1}^{11} T_n$,where the $n^{th}$ term $T_n$ is given by:$T_n = \frac{2n+1}{1^2 + 2^2 + \dots + n^2}$Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$T_n = \frac{2n+1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)}$Using partial fractions,we can write:$T_n = 6 \left( \frac{1}{n} - \frac{1}{n+1} ight)$Now,the sum of $n$ terms $S_n$ is:$S_n = \sum_{k=1}^n 6 \left( \frac{1}{k} - \frac{1}{k+1} ight) = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight]$$S_n = 6 \left( 1 - \frac{1}{n+1} ight) = \frac{6n}{n+1}$For $n=11$ terms:$S_{11} = \frac{6 	imes 11}{11+1} = \frac{66}{12} = \frac{11}{2}$

The sum $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ up to $11$ terms is

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