Question

Find the sum to $n$ terms of the sequence ${\left(x+\frac{1}{x}\right)}^2,{\left(x^2+\frac{1}{x^2}\right)}^2,{\left(x^3+\frac{1}{x^3}\right)}^2,\dots$.

Answer 1

Step1: Calculation of an equation for the sum of the given series.

Let $S_n$ denote the sum to $n$ terms of the given sequence.

$$\begin{gathered} S_n={\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+{\left(x^3+\frac{1}{x^3}\right)}^2+\dots +{\left(x^n+\frac{1}{x^n}\right)}^2 \\ \Rightarrow S_n=\left(x^2+\frac{1}{x^2}+2\right)+\left(x^4+\frac{1}{x^4}+2\right)+\left(x^6+\frac{1}{x^6}+2\right)+\dots +\left(x^{2n}+\frac{1}{x^{2n}}+2\right)\left(U\sin g{(a+b)}^2=a^2+b^2+2ab\right) \\ \therefore S_n=\left(x^2+x^4+x^6+\dots +x^{2n}\right)+\left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)+\left(2+2+2+\dots uptontimes\right)equation\left(1\right) \end{gathered}$$

Step2: Calculation of an expression $\left(x^2+x^4+x^6+\dots +x^{2n}\right)$.

The expression $\left(x^2+x^4+x^6+\dots +x^{2n}\right)$ is a G.P. series having first term $x^2$ and common ratio $x^2$.

The sum of the $n$ terms of a series G.P. is expressed as $S_n=a\left(\frac{r^n-1}{r-1}\right)$, where $a$ is the first term and $r$ is the common ratio.

Then, the sum of this series is given by:

$$\begin{gathered} \left(x^2+x^4+x^6+\dots +x^{2n}\right)=x^2\left(\frac{{\left(x^2\right)}^n-1}{x^2-1}\right) \\ \therefore \left(x^2+x^4+x^6+\dots +x^{2n}\right)=x^2\left(\frac{x^{2n}-1}{x^2-1}\right)equation\left(2\right) \end{gathered}$$

Step3: Calculation of an expression $\left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)$.

The expression $\left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)$ is a G.P. series having first term $\frac{1}{x^2}$ and common ratio $\frac{1}{x^2}$.

Then, the sum of this series is given by:

$$\begin{gathered} \left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)=\left(\frac{1}{x^2}\right)\left(\frac{{\left(\frac{1}{x^2}\right)}^n-1}{\frac{1}{x^2}-1}\right) \\ \Rightarrow \left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)=\left(\frac{1}{x^{2n}}\right)\left(\frac{1-x^{2n}}{1-x^2}\right) \\ \therefore \left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)=\left(\frac{1}{x^{2n}}\right)\left(\frac{x^{2n}-1}{x^2-1}\right)equation\left(3\right) \end{gathered}$$

Step4: Calculation of the sum of the series.

Substitute the values of $\left(x^2+x^4+x^6+\dots +x^{2n}\right)$ and $\left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\dots +\frac{1}{x^{2n}}\right)$ from equations (2) and (3) in equation (1).

$$\begin{gathered} S_n=x^2\left(\frac{x^{2n}-1}{x^2-1}\right)+\frac{1}{x^{2n}}\left(\frac{x^{2n}-1}{x^2-1}\right)+(2+2+2+\dots uptonterms) \\ \therefore S_n=\left(\frac{x^{2n}-1}{x^2-1}\right)\left(x^2+\frac{1}{x^{2n}}\right)+2n \end{gathered}$$

Final Answer: The sum to $n$ terms of the series ${\left(x+\frac{1}{x}\right)}^2,{\left(x^2+\frac{1}{x^2}\right)}^2,{\left(x^3+\frac{1}{x^3}\right)}^2,\dots$ is $\left(\frac{x^{2n}-1}{x^2-1}\right)\left(x^2+\frac{1}{x^{2n}}\right)+2n$.