Question

The sum of $10$ terms of the series ${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+\dots .$ is

• A. $\left(\frac{x^{20}-1}{x^2-1}\right)\left(\frac{x^{22}+1}{x^{20}}\right)+20$
• B. $\left(\frac{x^{18}-1}{x^2-1}\right)\left(\frac{x^{11}+1}{x^9}\right)+20$
• C. $\left(\frac{x^{18}-1}{x^2-1}\right)\left(\frac{x^{11}-1}{x^9}\right)+20$
• D. None of these

Answer 1

The correct option is A $\left(\frac{x^{20}-1}{x^2-1}\right)\left(\frac{x^{22}+1}{x^{20}}\right)+20$

$\begin{array}{cc}& {(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+\cdots \cdot +{(x^{10}+\frac{1}{x^{10}})}^2\\ \Rightarrow & (x^4+\frac{1}{x^4}+2)+(x^2+\frac{1}{x^2}+2)+\cdots +(x^{20}+\frac{1}{x^{2}0}+2)\\ \Rightarrow & (x^2+x^4+\cdots +x^{20})+(\frac{1}{x^2}+\frac{1}{x^4}+\cdots +\frac{1}{x^{2}0})+\left(2X10\right)\end{array}$

$\begin{array}{c}\Rightarrow x^2[\frac{{\left(x^2\right)}^{10}-1]}{x^2-1}+\frac{1}{x^2}\frac{[1-{\left(\frac{1}{x^2}\right)}^{10}]}{1-\frac{1}{x^2}}+20\\ \Rightarrow \frac{x^2(x^{20}-1)}{x^2-1}+\frac{1}{x^2}[\frac{\frac{x^{20}-1}{x^{20}}]}{\frac{x^2-1}{x^2}}+20\\ \Rightarrow \frac{x^2(x^{20}-1)}{x^2-1}+\frac{x^{20}-1}{x^{20}(x^2-1)}+20\\ \Rightarrow \frac{x^{20}-1}{x^2-1}[x^2+\frac{1}{x^{20}}]+20\end{array}$

$\begin{array}{c}\Rightarrow \left(\frac{x^{20}-1}{x^2-1}\right)\left(\frac{x^{22}+1}{x^{20}}\right)+20\\ \text{hence}\left(A\right)\text{is the correct option}\end{array}$