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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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,$\sum_{k=1}^r k^2 = \f

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,$\sum_{k=1}^r k^2 = \frac{r(r+1)(2r+1)}{6}$,we get:$T_r = \frac{2r+1}{\frac{r(r+1)(2r+1)}{6}} = \frac{6(2r+1)}{r(r+1)(2r+1)} = \frac{6}{r(r+1)}$.We can write this using partial fractions as $T_r = 6 \left( \frac{1}{r} - \frac{1}{r+1} ight)$.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:

Similar Questions

If $S$ is the sum of the first $10$ terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$ then $\tan ( S )$ is equal to

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

What is the sum of the first $n$ terms of the series whose $k$-th term is $k! \times k$?

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ to $n$ terms is

If $a_n = \frac{-2}{4n^2 - 16n + 15}$,then $a_1 + a_2 + \dots + a_{25}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module