Question

if $1+x^2=\sqrt{3}x,then,\prod _{n=1}^{24}\left(x^n+\frac{1}{x^n}\right)$ is equal to:

Answer 1

Given,

$1+x^2=\sqrt{3}x$

or, $x+\frac{1}{x}=\sqrt{3}$......(1).

Now,

$x^3+\frac{1}{x^3}$

$={(x+\frac{1}{x})}^3-3.x.\frac{1}{x}.(x+\frac{1}{x})$

$=3\sqrt{3}-3\sqrt{3}$ [ Using (1)]

$=0$.......(2).

Now,

$\prod _{n=1}^{24}\left(x^n+\frac{1}{x^n}\right)$

$=(x+\frac{1}{x}).(x^2+\frac{1}{x^2}).(x^3+\frac{1}{x^3})......(x^{24}+\frac{1}{x^{24}})$

$=0$. [ Using (2)]