Question

The value of the series \[{{\left( x+\dfrac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}+{{\left( x+\dfrac{1}{{{x}^{3}}} \right)}^{2}}...\] up to \[n\] terms is
A. \[\dfrac{{{x}^{2n}}-1}{{{x}^{2}}-1}\times \dfrac{{{x}^{2n+2}}+1}{{{x}^{2n}}}+2n\]
B. \[\dfrac{{{x}^{2n}}+1}{{{x}^{2}}+1}\times \dfrac{{{x}^{2n+2}}-1}{{{x}^{2n}}}-2n\]
C. \[\dfrac{{{x}^{2n}}-1}{{{x}^{2}}-1}\times \dfrac{{{x}^{2n+2}}-1}{{{x}^{2n}}}-2n\]
D. None of these

Answer 1

Hint: In this question, we have to find the sum of the given series. For, this the type of the series is to be known. By splitting the given series, we get two different series of the same type. Once the type of the series is known, we can easily calculate their sums. By simplifying them, we get the required value.

Formula Used: If the series is a geometric series, then the sum of the $n$ terms is calculated by
 ${{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ - First term in the series; $r$ - is Common ratio.
Some of the important formulae:
$\begin{align}
  & {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
 & {{(x+\dfrac{1}{x})}^{2}}={{x}^{2}}+2+\dfrac{1}{{{x}^{2}}} \\
 & {{(x-\dfrac{1}{x})}^{2}}={{x}^{2}}-2+\dfrac{1}{{{x}^{2}}} \\
\end{align}$

Complete step by step solution: Given series is
\[{{\left( x+\dfrac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}+{{\left( x+\dfrac{1}{{{x}^{3}}} \right)}^{2}}...\]
Then, consider the given series as
\[A=\left( {{x}^{2}}+{{x}^{4}}+{{x}^{6}}...n\text{ terms} \right)+\left( \dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}+\dfrac{1}{{{x}^{6}}}+...n\text{ terms} \right)+2n\text{ }...(1)\]
So, the sub-series formed in the main series are:
\[{{I}_{1}}={{x}^{2}}+{{x}^{4}}+{{x}^{6}}...n\text{ terms}\]
\[{{I}_{2}}=\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}+\dfrac{1}{{{x}^{6}}}+...n\text{ terms}\]
Then, their sums are calculated as follows:
\[\begin{align}
  & {{I}_{1}}={{x}^{2}}+{{x}^{4}}+{{x}^{6}}...n\text{ terms} \\
 & a={{x}^{2}};r={{x}^{2}}; \\
 & {{I}_{1}}=\dfrac{{{x}^{2}}({{x}^{2n}}-1)}{({{x}^{2}}-1)}\text{ }...(2) \\
\end{align}\]
And
\[\begin{align}
  & {{I}_{2}}=\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}+\dfrac{1}{{{x}^{6}}}+...n\text{ terms} \\
 & a=\dfrac{1}{{{x}^{2}}};r=\dfrac{1}{{{x}^{2}}} \\
 & {{I}_{2}}=\dfrac{\dfrac{1}{{{x}^{2}}}\left( 1-\dfrac{1}{{{x}^{2n}}} \right)}{1-\dfrac{1}{{{x}^{2}}}}\text{ }...(3) \\
\end{align}\]
Then, on substituting (2) and (3) in (1), we get
\[\begin{align}
  & A=\left( {{x}^{2}}+{{x}^{4}}+{{x}^{6}}...n\text{ terms} \right)+\left( \dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}+\dfrac{1}{{{x}^{6}}}+...n\text{ terms} \right)+2n \\
 & \text{ }=\dfrac{{{x}^{2}}({{x}^{2n}}-1)}{({{x}^{2}}-1)}+\dfrac{\dfrac{1}{{{x}^{2}}}\left( 1-\dfrac{1}{{{x}^{2n}}} \right)}{1-\dfrac{1}{{{x}^{2}}}}+2n \\
 & \text{ }=\dfrac{{{x}^{2}}({{x}^{2n}}-1)}{({{x}^{2}}-1)}+\dfrac{{{x}^{2n}}-1}{({{x}^{2}}-1){{x}^{2n}}}+2n \\
 & \text{ }=\dfrac{{{x}^{2n+2}}({{x}^{2n}}-1)+\left( {{x}^{2n}}-1 \right)}{{{x}^{2n}}({{x}^{2}}-1)}+2n \\
 & \text{ }=\dfrac{({{x}^{2n}}-1)}{({{x}^{2}}-1)}\times \dfrac{\left( {{x}^{2n+2}}+1 \right)}{{{x}^{2n}}}+2n \\
\end{align}\]

Option ‘A’ is correct

Note: Here we may forget to split the series into two other series. If we forget to split, the calculation may become difficult. To avoid that, we need to split the series into two and find their sum separately. Then, adding those sums, we get the required expression on simplifying.