The sum to 50 terms of the series frac{3}{1^2 | vedclass

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Question

(A) The $r$-th term of the series is given by $T_r = \frac{2r+1}{1^2 + 2^2 + \dots + r^2}$.Using the formula for the sum of squares of the first $r$ n

Vedclass · Accepted Answer

$\frac{100}{17}$

Vedclass · Answer

(A) The $r$-th term of the series is given by $T_r = \frac{2r+1}{1^2 + 2^2 + \dots + r^2}$.Using the formula for the sum of squares of the first $r$ natural numbers,$\sum_{k=1}^r k^2 = \frac{r(r+1)(2r+1)}{6}$,we get:$T_r = \frac{2r+1}{\frac{r(r+1)(2r+1)}{6}} = \frac{6(2r+1)}{r(r+1)(2r+1)} = \frac{6}{r(r+1)}$.Using partial fractions,$T_r = 6 \left( \frac{1}{r} - \frac{1}{r+1} ight)$.The sum of $n$ terms is $S_n = \sum_{r=1}^n T_r = 6 \sum_{r=1}^n \left( \frac{1}{r} - \frac{1}{r+1} ight)$.This is a telescoping series: $S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight] = 6 \left( 1 - \frac{1}{n+1} ight) = \frac{6n}{n+1}$.For $n = 50$,$S_{50} = \frac{6 	imes 50}{50+1} = \frac{300}{51} = \frac{100}{17}$.

The sum to $50$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ is:

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