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I am stumped by the sum $$\sum_{x=0}^n \binom{n}{x}\sin\left(\frac{\pi x}{n}\right)$$ but I can't figure it out. I tried expanding the taylor series of sine and using Euler's identity, but to no avail. Any hints?

PLEASE do not give me a full solution - I just need a hint.

Thanks!

Blue
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Franklin Pezzuti Dyer
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2 Answers2

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As you exclusively asked for an hint, here it is:

$$\sum_{x=0}^n\binom{n}x\sin\frac{\pi x}n=\Im\left(\sum_{x=0}^n\binom{n}xe^{\frac{\imath\pi x}n}\right)$$

Now, use the binomial theorem and the result $1+e^{\imath x}=2\cos\frac{x}2e^\frac{\imath x}2$ to arrive at the answer.

Integreek
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The hint is to use Euler’s form of any imaginary number. It clicked me as soon as I saw nCr, must use binomial theorem, would be useful if I get some expression of the form nCr times x^r, and euler’s constant fits everything! That’s for the hint, the final solution comes out to be:-

The function ( S(n) ) is defined as:

…S(n) =\begin{cases} 0 & \text{if } n \text{ is odd}, \\ 2^{n/2} \cdot \sin\left(\frac{n\pi}{4}\right) & \text{if } n \text{ is even}. \end{cases}

Would love any critiques, and let me know if further clarification is required.

Kushman
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