JEE Mains · Maths · STD 11 - 8. sequence and series
The \(\operatorname{sum} \sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}\) is equal to
- A \(\frac{7}{87}\)
- B \(\frac{7}{29}\)
- C \(\frac{14}{87}\)
- D \(\frac{21}{29}\)
Answer & Solution
Correct Answer
(B) \(\frac{7}{29}\)
Step-by-step Solution
Detailed explanation
\(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}=\frac{3}{4} \sum_{n=1}^{21} \frac{1}{4 n-1}-\frac{1}{4 n+3}\) \(=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\ldots .+\left(\frac{1}{83}-\frac{1}{87}\right)\right]\)…
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