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

Concept: The general term of the series can be written as: \[ T_r = \left(x^r + \frac{1}{x^r}\right)^2 = (x^r)^2 + \frac{1}{(x^r)^2} + 2(x^r)\left(\frac{1}{x^r}\right) = x^{2r} + \frac{1}{x^{2r}} + 2 \]

Step 1: Form the sum of $n$ terms. The sum $S_n$ is: \[ S_n = \sum_{r=1}^{n} \left( x^{2r} + \frac{1}{x^{2r}} + 2 \right) \] \[ S_n = \sum_{r=1}^{n} x^{2r} + \sum_{r=1}^{n} \frac{1}{x^{2r}} + \sum_{r=1}^{n} 2 \]

Step 2: Evaluate the individual summations. 1. The first term is a geometric progression (G.P.) with first term $a = x^2$ and common ratio $r = x^2$: \[ \sum_{r=1}^{n} x^{2r} = x^2 \frac{(x^2)^n - 1}{x^2 - 1} = \frac{x^2(x^{2n}-1)}{x^2-1} \] 2. The second term is a G.P. with $a = \frac{1}{x^2}$ and common ratio $r = \frac{1}{x^2}$: \[ \sum_{r=1}^{n} \frac{1}{x^{2r}} = \frac{1}{x^2} \frac{1 - (\frac{1}{x^2})^n}{1 - \frac{1}{x^2}} = \frac{1}{x^2} \frac{\frac{x^{2n}-1}{x^{2n}}}{\frac{x^2-1}{x^2}} = \frac{x^{2n}-1}{x^{2n}(x^2-1)} \] 3. The third term is simply: $\sum_{r=1}^{n} 2 = 2n$.

Step 3: Combine the results. \[ S_n = \frac{x^2(x^{2n}-1)}{x^2-1} + \frac{x^{2n}-1}{x^{2n}(x^2-1)} + 2n \] Factor out $\frac{x^{2n}-1}{x^2-1}$: \[ S_n = \frac{x^{2n}-1}{x^2-1} \left( x^2 + \frac{1}{x^{2n}} \right) + 2n = \frac{x^{2n}-1}{x^2-1} \left( \frac{x^{2n+2}+1}{x^{2n}} \right) + 2n \]