JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the first \(10\) terms of the series \(\dfrac{1}{1 + 1^4 \times 4} + \dfrac{2}{1 + 2^4 \times 4} + \dfrac{3}{1 + 3^4 \times 4} + \dfrac{4}{1 + 4^4 \times 4} + \ldots\) is \(\dfrac{m}{n}\), \(\gcd(m, n) = 1\), then \(m + n\) is equal to :
- A \(256\)
- B \(264\)
- C \(276\)
- D \(284\)
Answer & Solution
Correct Answer
(C) \(276\)
Step-by-step Solution
Detailed explanation
The general term of the series is \(T_r = \dfrac{r}{1 + 4r^4}\). Factorizing the denominator: \(1 + 4r^4 = (1 + 2r^2)^2 - 4r^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)\) Thus, \(T_r = \dfrac{r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}\) Multiplying and dividing by \(4\):…
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