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Problem 873

AIME late

Combinatorics Difficulty 5.6 Prove it

Example. Prove: (1) $C_{n}^{0}-C_{n}^{2}+C_{n}^{4}-C_{n}^{6}+\ldots$
$=2^{\frac{n}{2}} \cos \frac{n \pi}{4} \text {; }$
(2) $C_{n}^{1}-C_{n}^{3}+C_{n}^{5}-C_{n}^{7}+\cdots$
$= 2^{\frac{n}{2}} \sin \frac{n \pi}{4} .$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

The proof method is to choose $(1+\mathrm{i})^{n}=\left[\sqrt{2}\left(\cos \frac{\pi}{4}\right.\right.$ $\left.\left.+i \sin \frac{\pi}{4}\right)\right]^{n}=2^{\frac{n}{2}}\left(\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right)$ , compare the real part coefficients on both sides to get the first question, and compare the imaginary part coefficients on both sides to get the second question.
Using the same method, we can prove:
(1)
$\begin{array}{l} C_{n}^{1}-3 C_{n}^{s}+5 C_{n}^{s}-7 C_{n}^{1}+\cdots=n \cdot 2^{\frac{n-1}{2}} \\ \text { - } \cos \frac{n-1}{4} \pi ; \end{array}$
(2) 2
$\begin{array}{l} 2 C_{n}^{2}-4 C_{n}^{4}+6 C_{n}^{8}-8 C_{n}^{8}+\cdots=n \cdot 2^{\frac{n-1}{2}} \\ \cdot \sin \frac{n-1}{4} \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.