If the sum frac{3}{1^2} + frac{5}{1^2 + 2^2} | vedclass

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(A) The $n^{th}$ term of the given series is given by:$a_n = \frac{2n + 1}{1^2 + 2^2 + \dots + n^2}$Using the formula for the sum of squares of the fi

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$120$

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(A) The $n^{th}$ term of the given series is given by:$a_n = \frac{2n + 1}{1^2 + 2^2 + \dots + n^2}$Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$a_n = \frac{2n + 1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)}$We can use partial fractions to simplify $a_n$:$a_n = 6 \left( \frac{1}{n} - \frac{1}{n+1} \right)$Now,the sum of $20$ terms is $S_{20} = \sum_{n=1}^{20} a_n = 6 \sum_{n=1}^{20} \left( \frac{1}{n} - \frac{1}{n+1} \right)$This is a telescoping series:$S_{20} = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{20} - \frac{1}{21}) \right]$$S_{20} = 6 \left( 1 - \frac{1}{21} \right) = 6 \left( \frac{20}{21} \right) = \frac{120}{21}$Given that $S_{20} = \frac{k}{21}$,by comparing,we get $k = 120$.

If the sum $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ up to $20$ terms is equal to $\frac{k}{21}$,then $k$ is equal to

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