Maths Olympiad Prep

Track / Stage 5 / 274 of 400 #874 of 1964

Problem 874

AIME late
Algebra Difficulty 5.7 Find the answer

330. Simplify the sums:

a) ctg2π2n+1+ctg22π2n+1+ctg23π2n+1++ctg2nπ2n+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) cosec2π2n+1+cosec22π2n+1+cosec23π2n+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

+cosec2nπ2n+1\cdots+\operatorname{cosec}^{2} \frac{n \pi}{2 n+1}.

A number or a short expression. Spacing and $ signs are ignored.

Official solution

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

xnC2n+13C2n+11xn1+C2n+15C2n+11xn2=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 xn1x^{n-1}, taken with the opposite sign, i.e.

ctg2π2n+1+ctg22π2n+1+ctg23π2n+1++ctg2nπ2n+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(2n1)3 =\frac{C_{2 n+1}^{3}}{C_{2 n+1}^{1}}=\frac{n(2 n-1)}{3}

b) Since cosec2α=ctg2α+1\operatorname{cosec}^{2} \alpha=\operatorname{ctg}^{2} \alpha+1, it follows from the formula of problem a) that

cosec2π2n+1+cosec22π2n+1+cosec23π2n+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

+cosec2nπ2n+1=n(2n1)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 and solution reproduced as published; topic, difficulty and ordering added by this site.