If the sum $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ up to $20$ terms is equal to $\frac{k}{21}$,then $k$ is equal to

  • A
    $120$
  • B
    $180$
  • C
    $240$
  • D
    $60$

Explore More

Similar Questions

$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}} \cdot r! = k!$,then the number of divisors of $k$ is

$\sum_{r=1}^{20} (r^{2}+1)(r!)$ is equal to:

The sum of the first $n$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ is $.........$.

Difficult
View Solution

If $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d$,then the sum of the series $\sin d (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n)$ is:

The value of $1000 \left[ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{999 \times 1000} \right]$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo