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 the sum of the first $20$ terms of the series $\frac{4.1}{4+3.1^2+1^4}+\frac{4.2}{4+3.2^2+2^4}+\frac{4.3}{4+3.3^2+3^4}+\frac{4.4}{4+3.4^2+4^4}+\ldots$ is $\frac{m}{n}$,where $m$ and $n$ are coprime,then $m + n$ is equal to :-

The sum $1 \times 1! + 2 \times 2! + \ldots + 50 \times 50!$ equals

$\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \ldots$ to $50$ terms $=$

If ${a_k} = \frac{1}{{k(k + 1)}}$ for $k = 1, 2, 3, 4, ..., n$,then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

$\frac{1}{3 \cdot 6} + \frac{1}{6 \cdot 9} + \frac{1}{9 \cdot 12} + \dots$ to $9$ terms $=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module