JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the first ten terms of the series \(\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\ldots\) is \(\frac{m}{n}\), where \(m\) and \(n\) are co-prime numbers, then \(m + n\) is equal to
- A \(280\)
- B \(277\)
- C \(276\)
- D \(272\)
Answer & Solution
Correct Answer
(C) \(276\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\ldots \ldots\) \(T_{n}=\frac{n}{4 n^{4}+1}\) \(=\frac{ n }{\left(2 n^{2}+1\right)^{2}-(2 n )^{2}}=\frac{ n }{\left(2 n ^{2}+2 n +1\right)\left(2 n ^{2}-2 n +1\right)}\)…
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