Question

Find the sum of the series \( \left(x + \frac{1}{x}\right)^2 + \left(x^2 + \frac{1}{x^2}\right)^2 + \left(x^3 + \frac{1}{x^3}\right)^2 + \cdots \) up to \( n \) terms.

• A. \( \frac{x^{2n}-1}{x^2-1} \times \frac{x^{2n+2}+1}{x^{2n}} + 2n \)
• B. \( \frac{x^{2n}+1}{x^2+1} \times \frac{x^{2n+2}-1}{x^{2n}} - 2n \)
• C. \( \frac{x^{2n}-1}{x^2-1} \times \frac{x^{2n}-1}{x^{2n}} - 2n \)
• D. \( \text{None of these} \)

Hint:

Expand the square first to simplify the expression into separate geometric series that are easier to sum.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
This problem asks us to find the sum of a series where each term is the square of a sum containing a variable \( x \) and its reciprocal raised to consecutive integer powers. We need to sum this series up to \( n \) terms.
Step 2 : Key Formulas and Approach:
The general term of the series can be expanded algebraically using the identity:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
The expanded terms can then be grouped into separate geometric progressions (G.P.). The formula for the sum of the first \( n \) terms of a G.P. is:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
where \( a \) is the first term and \( r \) is the common ratio.
Step 3 : Detailed Explanation:

Write down the general term \( T_r \) of the given series:
\[ T_r = \left(x^r + \frac{1}{x^r}\right)^2 \]

Expand \( T_r \) using the algebraic square identity:
\[ T_r = (x^r)^2 + \left(\frac{1}{x^r}\right)^2 + 2(x^r)\left(\frac{1}{x^r}\right) = x^{2r} + \frac{1}{x^{2r}} + 2 \]

Express the total sum of \( n \) terms \( S_n \) by summing \( T_r \) from \( r = 1 \) to \( n \):
\[ S_n = \sum_{r=1}^{n} \left( x^{2r} + \frac{1}{x^{2r}} + 2 \right) \]

Split the summation into three individual parts:
\[ S_n = \sum_{r=1}^{n} x^{2r} + \sum_{r=1}^{n} \frac{1}{x^{2r}} + \sum_{r=1}^{n} 2 \]

Evaluate the first summation, which is a G.P. with first term \( a = x^2 \) and common ratio \( r = x^2 \):
\[ \sum_{r=1}^{n} x^{2r} = \frac{x^2(x^{2n} - 1)}{x^2 - 1} \]

Evaluate the second summation, which is a G.P. with first term \( a = \frac{1}{x^2} \) and common ratio \( r = \frac{1}{x^2} \):
\[ \sum_{r=1}^{n} \frac{1}{x^{2r}} = \frac{1}{x^2} \frac{1 - (\frac{1}{x^2})^n}{1 - \frac{1}{x^2}} = \frac{x^{2n} - 1}{x^{2n}(x^2 - 1)} \]

Evaluate the third constant summation: \( \sum_{r=1}^{n} 2 = 2n \).

Combine the three individual sums together:
\[ S_n = \frac{x^2(x^{2n} - 1)}{x^2 - 1} + \frac{x^{2n} - 1}{x^{2n}(x^2 - 1)} + 2n \]

Factor out the common term \( \frac{x^{2n} - 1}{x^2 - 1} \) to simplify:
\[ S_n = \frac{x^{2n}-1}{x^2-1} \left( x^2 + \frac{1}{x^{2n}} \right) + 2n = \frac{x^{2n}-1}{x^2-1} \left( \frac{x^{2n+2}+1}{x^{2n}} \right) + 2n \]

Step 4 : Final Answer:
The sum of the series is \( \frac{x^{2n}-1}{x^2-1} \times \frac{x^{2n+2}+1}{x^{2n}} + 2n \times \), which corresponds to option (A).