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∑n=1∞4nsin4(π2n)\large\displaystyle\sum_{n=1}^{\infty}4^n\sin^4\left(\dfrac{\pi}{2^n}\right)n=1∑∞4nsin4(2nπ)
If the value of the summation above is in the form of πab\dfrac{\pi^a}bbπa, where aaa and bbb are positive integers, find the value of a+ba+ba+b.
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