The sum $\displaystyle\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}$ is equal to
- $\frac{7}{87}$
- $\frac{7}{29}$
- $\frac{14}{87}$
- $\frac{21}{29}$
The Correct Option is B
Approach Solution - 1
So, The correct option is (B): $\frac7{29}$
Approach Solution -2
\(\overset{21}{\underset{n=1}\sum} \frac3{(4n-1)(4n+3)} = \frac3{4}\overset{21}{\underset{n=1}\sum} \frac1{(4n-1)}-\frac1{(4n+3)}\)
=43n=1∑21(4n−1)(4n+3)(4n+3)−(4n−1)
=43n=1∑214n−11−4n+31
=43(31−71+71−111+111−….+831−871)
=43(31−871)=297
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Concepts Used:
Series
A collection of numbers that is presented as the sum of the numbers in a stated order is called a series. As an outcome, every two numbers in a series are separated by the addition (+) sign. The order of the elements in the series really doesn't matters. If a series demonstrates a finite sequence, it is said to be finite, and if it demonstrates an endless sequence, it is said to be infinite.
Read More: Sequence and Series
Types of Series:
The following are the two main types of series are: