The value of sumlimits_{k = 1}^infty {frac{{ | vedclass

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

Question

(D) We observe that the numerator can be written as $(k+1)^3 - k^3 = k^3 + 3k^2 + 3k + 1 - k^3 = 3k^2 + 3k + 1$. Thus,the general term is $\frac{(k+1)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We observe that the numerator can be written as $(k+1)^3 - k^3 = k^3 + 3k^2 + 3k + 1 - k^3 = 3k^2 + 3k + 1$. Thus,the general term is $\frac{(k+1)^3 - k^3}{k^3(k+1)^3} = \frac{(k+1)^3}{k^3(k+1)^3} - \frac{k^3}{k^3(k+1)^3} = \frac{1}{k^3} - \frac{1}{(k+1)^3}$. This is a telescoping series. The sum is $\sum_{k=1}^n \left( \frac{1}{k^3} - \frac{1}{(k+1)^3} ight) = \left( 1 - \frac{1}{2^3} ight) + \left( \frac{1}{2^3} - \frac{1}{3^3} ight) + \dots + \left( \frac{1}{n^3} - \frac{1}{(n+1)^3} ight)$. As $n 	o \infty$,the sum converges to $1 - 0 = 1$.

The value of $\sum\limits_{k = 1}^\infty {\frac{{3{k^2} + 3k + 1}}{{{{\left( {{k^2} + k} \right)}^3}}}} $ is equal to

Similar Questions

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

The sum $\sum\limits_{r = 1}^{10} {({r^2} + 1) \times r!}$ is equal to

The sum of the infinite series ${\tan ^{ - 1}}\left( {\frac{2}{{1 - {1^2} + {1^4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{4}{{1 - {2^2} + {2^4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{6}{{1 - {3^2} + {3^4}}}} \right) + \dots$ is

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

If $\frac{1}{2 \times 7} + \frac{1}{7 \times 12} + \frac{1}{12 \times 17} + \frac{1}{17 \times 22} + \dots$ to $10$ terms $= k$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module