Maths Olympiad Prep

Track / Stage 5 / 333 of 400 #933 of 1964

Problem 933

AIME late
Algebra Difficulty 5.7 Find the answer

Simplify the sums:

a) ctg⁡2π2n+1+ctg⁡22π2n+1+ctg⁡23π2n+1+…+ctg⁡2nπ2n+1\operatorname{ctg}^{2} \frac{\pi}{2 n+1}+\operatorname{ctg}^{2} \frac{2 \pi}{2 n+1}+\operatorname{ctg}^{2} \frac{3 \pi}{2 n+1}+\ldots+\operatorname{ctg}^{2} \frac{n \pi}{2 n+1};

b) cosec⁡2π2n+1+cosec⁡22π2n+1+cosec⁡23π2n+1+…\operatorname{cosec}^{2} \frac{\pi}{2 n+1}+\operatorname{cosec}^{2} \frac{2 \pi}{2 n+1}+\operatorname{cosec}^{2} \frac{3 \pi}{2 n+1}+\ldots

⋯+cosec⁡2nπ2n+1\cdots+\operatorname{cosec}^{2} \frac{n \pi}{2 n+1}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

330. a) The sum of the roots of the equation of degree nn

xn−C2n+13C2n+11xn−1+C2n+15C2n+11xn−2−…=0 x^{n}-\frac{C_{2 n+1}^{3}}{C_{2 n+1}^{1}} x^{n-1}+\frac{C_{2 n+1}^{5}}{C_{2 n+1}^{1}} x^{n-2}-\ldots=0

(see the solution of problem 229 b)) is equal to the coefficient of xn−1x^{n-1}, taken with the opposite sign, i.e.

ctg⁡2π2n+1+ctg⁡22π2n+1+ctg⁡23π2n+1+…+ctg⁡2nπ2n+1=\operatorname{ctg}^{2} \frac{\pi}{2 n+1}+\operatorname{ctg}^{2} \frac{2 \pi}{2 n+1}+\operatorname{ctg}^{2} \frac{3 \pi}{2 n+1}+\ldots+\operatorname{ctg}^{2} \frac{n \pi}{2 n+1}=

=C2n+13C2n+11=n(2n−1)3 =\frac{C_{2 n+1}^{3}}{C_{2 n+1}^{1}}=\frac{n(2 n-1)}{3}

b) Since cosec⁡2α=ctg⁡2α+1\operatorname{cosec}^{2} \alpha=\operatorname{ctg}^{2} \alpha+1, it follows from the formula of problem a) that

cosec⁡2π2n+1+cosec⁡22π2n+1+cosec⁡23π2n+1+…\operatorname{cosec}^{2} \frac{\pi}{2 n+1}+\operatorname{cosec}^{2} \frac{2 \pi}{2 n+1}+\operatorname{cosec}^{2} \frac{3 \pi}{2 n+1}+\ldots

…+cosec⁡2nπ2n+1=n(2n−1)3+n=2n(n+1)3 \ldots+\operatorname{cosec}^{2} \frac{n \pi}{2 n+1}=\frac{n(2 n-1)}{3}+n=\frac{2 n(n+1)}{3}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.