# Evaluating $prod^{100}_{k=1}left[1+2cos frac{2pi cdot 3^k}{3^{100}+1}right]$

Mathematics Asked by jacky on December 13, 2020

Evaluate$$prod^{100}_{k=1}left[1+2cos frac{2pi cdot 3^k}{3^{100}+1}right]$$

My attempt:

$$1+2cos 2theta= 1+2(1-2sin^2theta)=3-4sin^2theta$$

$$=frac{3sin theta-4sin^3theta}{sin theta}=frac{sin 3theta}{sin theta}$$

I did not understand how to solve after that. Help required.

It's unclear if you are asking about $$prod^{100}_{k=1}left[1+2cos frac{2pi cdot 3^k}{3^{100}+1}right]$$ or about $$prod^{100}_{k=1}left[1+2cos frac{2pi kcdot 3^k}{3^{100}+1}right]$$

I will assume it is the former. At the moment, that is in the body of your question, while the latter is in the title. If it's the former, then using the trig identity you found, you have a telescoping product which then further simplifies nicely: begin{align} prod^{100}_{k=1}frac{sinleft(pifrac{3^{k+1}}{3^{100}+1}right)}{sinleft(pifrac{3^{k}}{3^{100}+1}right)}&=frac{sinleft(pifrac{3^{2}}{3^{100}+1}right)}{sinleft(pifrac{3^{1}}{3^{100}+1}right)}frac{sinleft(pifrac{3^{3}}{3^{100}+1}right)}{sinleft(pifrac{3^{2}}{3^{100}+1}right)}cdotsfrac{sinleft(pifrac{3^{101}}{3^{100}+1}right)}{sinleft(pifrac{3^{100}}{3^{100}+1}right)}\[1pc] &=frac{sinleft(pifrac{3^{101}}{3^{100}+1}right)}{sinleft(pifrac{3^{1}}{3^{100}+1}right)}\[1pc] &=frac{sinleft(pifrac{3cdot3^{100}}{3^{100}+1}right)}{sinleft(pifrac{3}{3^{100}+1}right)}\[1pc] &=frac{sinleft(pifrac{3left(cdot3^{100}+1right)-3}{3^{100}+1}right)}{sinleft(pifrac{3}{3^{100}+1}right)}\[1pc] &=frac{sinleft(3pi-pifrac{3}{3^{100}+1}right)}{sinleft(pifrac{3}{3^{100}+1}right)}\[1pc] &=frac{sinleft(pifrac{3}{3^{100}+1}right)}{sinleft(pifrac{3}{3^{100}+1}right)}\[1pc] &=1 end{align}

Note near the end that $sin(3pi-X)=sin(X)$.

Correct answer by alex.jordan on December 13, 2020

Let $$z=cosbigg(frac{2pi}{3^n+1}bigg)+isinbigg(frac{2pi}{3^n+1}bigg)$$

Then $z^{3^n+1}=1$ and also $displaystyle 2cos bigg(frac{2picdot 3^k}{3^n+1}bigg)=z^{3^k}+frac{1}{z^{3^k}}$

Write $$prod^{n}_{k=1}bigg[1+2cosbigg(frac{2picdot 3^k}{3^n+1}bigg)bigg]$$ $$=bigg(1+z^3+frac{1}{z^3}bigg)bigg(1+z^9+frac{1}{z^9}bigg)cdots cdots bigg(1+z^{3n}+frac{1}{z^{3n}}bigg)$$ $$=frac{(1+z^3+z^6)(1+z^9+z^{18})cdots cdots (1+z^{3^n}+(z^{3^n})^2)}{z^{3+9+cdots cdots +3^n}}$$

Multiply both Nr and Dr by $(1-z^3)$

$$=frac{1-z^{3^{n+1}}}{(1-z^3)cdot z^{3frac{(3^n-1)}{2}}}= frac{1-z^{-3}}{-(1-z^3)cdot z^{-3}}=1.$$

$text{Simplification}:;;$ From $z^{3n+1}=1Rightarrow z^{3n}=z^{-1}$

And $$z^{3frac{(3^n-1)}{2}}=z^{-3}cdot z^{3frac{(3^+1)}{2}}=z^{-3}cdot bigg(z^{frac{(3^+1)}{2}}bigg)^3=-z^3$$

Answered by jacky on December 13, 2020