If frac{1}{2 times 4} + frac{1}{4 times 6} + | vedclass

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

Question

(A) The given series is $S_n = \sum_{r=1}^{n} \frac{1}{(2r)(2r+2)}$.We can write the general term as $\frac{1}{4} 	imes \frac{1}{r(r+1)} = \frac{1}{4

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) The given series is $S_n = \sum_{r=1}^{n} \frac{1}{(2r)(2r+2)}$.We can write the general term as $\frac{1}{4} 	imes \frac{1}{r(r+1)} = \frac{1}{4} \left( \frac{1}{r} - \frac{1}{r+1} ight)$.Summing this from $r=1$ to $n$:$S_n = \frac{1}{4} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight]$.This is a telescoping series,so $S_n = \frac{1}{4} \left( 1 - \frac{1}{n+1} ight)$.$S_n = \frac{1}{4} \left( \frac{n+1-1}{n+1} ight) = \frac{n}{4(n+1)}$.Comparing this with $\frac{k n}{n+1}$,we get $k = \frac{1}{4}$.

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to

Similar Questions

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} = $

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$, then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module