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How do I show that $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\cos2kx =2^n(n+1)\cos^nx\cos(n+2)x\tag{1}$$ $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\sin2kx =2^n(n+1)\cos^nx\sin(n+2)x\tag{2}$$

My try:

From power-reduction formula, if $n$ is odd, we have $$\cos^{n}x = \frac{2}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos(n-2k)x\tag{3}$$ Putting (3) in (1), assuming $n$ is odd in (1), we have $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\cos2kx =2^{n}(n+1)\cos(n+2)x\frac{2}{2^n}\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos(n-2k)x$$ $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\cos2kx =2(n+1)\cos(n+2)x\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos(n-2k)x$$ $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\cos2kx =2(n+1)\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\cos(n+2)x\cos(n-2k)x \tag{4}$$ Using the product-to-sum identity on the RHS of (4), we have $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\cos2kx =2(n+1)\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\left(\frac{\cos2(n-k+1)x+\cos2(k+1)x}{2}\right)$$ $$\sum_{k=0}^{n+1}\binom{n+1}{k}k\cos2kx =(n+1)\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k}\left(\cos2(n-k+1)x+\cos2(k+1)x\right)$$

This is where I got stuck. I am thinking if the odd part could be solved, it will pave the way for the even part. Same for (2).

How can I proceed or is there a simpler way of solving it?

Blue
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Shuaib Lateef
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1 Answers1

2

Hint

Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

$$C+iS$$

$$=\sum_{k=0}^{n+1}\binom{n+1}kk(e^{2ix})^k$$

$$=(n+1)e^{2ix}\sum_{k=1}^{n+1}\binom n{k-1}(e^{2ix})^{k-1}$$

$$=(n+1)e^{2ix}(1+e^{2ix})^n$$

as for $k>0,k\binom{n+1}k=(n+1)\binom n{k-1}$

Now $1+e^{2ix}=e^{ix}(e^{ix}+e^{-ix})=e^{ix}(2\cos x)$

lab bhattacharjee
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