Possible telescopic sum

Question

Prove that $$\sum_{k=1}^n 4^{k}\sin^{4} \left(\frac{a}{2^k}\right) = 4^{n}\sin^{2} \left(\frac{a}{2^n}\right) - \sin^{2}a$$

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I suspect that telescopic sum is involved but don't know how to proceed. Please do not use induction, since the problem was given as objective.

Please help me with this problem.

Answer 1

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$4^k\sin^4(\dfrac{a}{2^k})=4^k\sin^2(\dfrac{a}{2^k})\color{blue}{\sin^2(\dfrac{a}{2^k})}=4^k\sin^2(\dfrac{a}{2^k})-4^k\sin^2(\dfrac{a}{2^k})\cos^2(\dfrac{a}{2^k})$

And then, with the double angle formula $\sin 2x= 2\sin x\cos x$:

$\sin^2(\dfrac{a}{2^k})\cos^2(\dfrac{a}{2^k})=(4^{-1})\sin^2(\dfrac{a}{2^{k-1}})$

Substitute the right side of the second equation for the sine-cosine product in the first one and identify the telescoping difference.

Answer 2

$$\sin^4 x= \sin^2 x(1-\cos^2 )= \sin^2 x- 4^{-1} \sin^2 2x$$ Let $x=a/2^n$ then we get the right differencing as

$$4^n\sin^4\frac{ a}{2^n}= \left(4^n \sin^2 \frac{a}{2^n}- 4^{n-1} \sin^2 \frac{a}{2^{n-1}} \right),$$ For telescopic summation.