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Prove that $$ \\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} ( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} $$ for all values of $x$.

I came across this joke which said that you could just bring the powers out of the brackets and everything works out right.

$$ \require{cancel} \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} ( \sin^{2(\textbf{n+r+1})}x + \cos^{2(\textbf{n+r+1})}x )\right) \\ = \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} (\cancelto1{\sin^2x + \cos^2x })^{\textbf{n+r+1}}\right) \\ = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} $$

But how would you actually go about proving this? I noticed that this value is also equal to $B(n+1,n+1)$, which may or may not be relevant.

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Let's prove $$ \\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} ( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} \tag1 $$ for all values of $x$.

By differentiating both sides with respect to $x$, on the right hand side one gets $0$, on the left hand side one gets $$ \sum_{r=0}^n (-1)^r {n \choose r} \left( 2\cos x\cdot\sin^{2(n+r+1)-1}x - 2\sin x\cdot\cos^{2(n+r+1)-1}x \right) $$ $$ 2\cos x\cdot\sin^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r} \sin^{2r}x - 2\sin x\cdot\cos^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r} \cos^{2r}x $$ $$ 2\cos x\cdot\sin^{2n+1}x\cdot \left(1-\sin^2x \right)^n-2\sin x\cdot\cos^{2n+1}x\cdot \left(1-\cos^2x \right)^n=0. $$ Since both sides of $(1)$ clearly agree at $x=0$, then $(1)$ is true for all values of $x$.

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