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Q.The sum of the series k=16k(2k1)(2k+1)4k\sum_{k=1}^{\infty} \dfrac{6^k}{(2k-1)(2k+1) \cdot 4^k} is

a
1π41 - \dfrac{\pi}{4}
b
π412\dfrac{\pi}{4} - \dfrac{1}{2}
c
π4\dfrac{\pi}{4}
d
11

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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