Question

Calculate the sum of the series ${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+.....{(x^n+\frac{1}{x^n})}^2$

• A. $\frac{x^n-1}{x^2-1}\cdot \left[\frac{x^{2n+2}+1}{x^{2n}}\right]+2n.$
• B. $\frac{x^n-1}{x^2-1}\cdot \left[\frac{x^{n+1}+1}{x^{2n}}\right]+2n.$
• C. $\frac{x^{2n}-1}{x^2-1}\cdot \left[\frac{x^{2n+2}+1}{x^{2n}}\right]+2n.$
• D. $\frac{x^{2n}-1}{x^2-1}\cdot \left[\frac{x^{n+1}+1}{x^{2n}}\right]+2n.$

Answer 1

Answer: (C)

$\frac{x^{2n}-1}{x^2-1}\cdot \left[\frac{x^{2n+2}+1}{x^{2n}}\right]+2n.$

$S=$ ${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+.....{(x^n+\frac{1}{x^n})}^2$
Open the brackets and splitting into three series
$S=$ $(x^2+x^4+x^6+...)+(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+...)+(2+2+2+...)$
$=x^2\frac{(x^{2n}-1)}{x^2-1}+\frac{1}{x^2}\frac{\left(\frac{1}{x^{2n}}-1\right)}{\frac{1}{x^2}-1}+2n$
$=\frac{x^2(x^{2n}-1)}{x^2-1}+\frac{x^{2n}-1}{x^{2n}.(x^2-1)}+2n=\frac{x^{2n}-1}{x^2-1}[x^2+\frac{1}{x^{2n}}]+2n=\frac{x^{2n}-1}{x^2-1}\cdot \left[\frac{x^{2n+2}+1}{x^{2n}}\right]+2n$