Question

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

• 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. None of these

Hint:

When resolving complex progression formula matching, use the Substitution Strategy ($n = 1$) to bypass long algebraic layouts entirely! For \(n = 1\), the true sum is simply the very first term: \(\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2\). Let's substitute \(n = 1\) directly into Option 1: \[ \frac{x^2 - 1}{x^2 - 1} \times \frac{x^{2+2} + 1}{x^2} + 2(1) = 1 \times \left(\frac{x^4 + 1}{x^2}\right) + 2 = x^2 + \frac{1}{x^2} + 2 \] It matches perfectly in less than 15 seconds!

Answer 1

The Correct Option is A

Concept: This problem can be solved analytically by expanding each term algebraically and grouping them into independent Geometric Progressions (G.P.). The standard sum formula for a geometric progression consisting of \(n\) terms with a first term \(a\) and common ratio \(r\) is given by: \[ S_n = \frac{a(r^n - 1)}{r - 1} \]

Step 1: Expanding the terms and splitting the series.
Let \(S\) denote the sum of the series. Expanding each term using the identity \((A+B)^2 = A^2 + 2AB + B^2\): (x + \frac{1}{x})^2 &= x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = x^2 + \frac{1}{x^2} + 2
(x^2 + \frac{1}{x^2})^2 &= x^4 + 2(x^2)(\frac{1}{x^2}) + \frac{1}{x^4} = x^4 + \frac{1}{x^4} + 2
(x^3 + \frac{1}{x^3})^2 &= x^6 + 2(x^3)(\frac{1}{x^3}) + \frac{1}{x^6} = x^6 + \frac{1}{x^6} + 2 Summing these expanded expressions together up to \(n\) terms: \[ S = \left(x^2 + \frac{1}{x^2} + 2\right) + \left(x^4 + \frac{1}{x^4} + 2\right) + \left(x^6 + \frac{1}{x^6} + 2\right) + \dots \text{ up to } n \text{ blocks} \] Regrouping the expressions into three separate summation series: S = {} & \underbrace{(x^2 + x^4 + \dots + x^{2n})}_{Series 1 (G.P.)}
& + \underbrace{(\frac{1}{x^2} + \frac{1}{x^4} + \dots + \frac{1}{x^{2n}})}_{Series 2 (G.P.)}
& + \underbrace{(2 + 2 + \dots + 2)}_{n times} \nobreak

Step 2: Evaluating each individual sub-series.

• Series 1: This is a G.P. with first term \(a = x^2\) and common ratio \(r = x^2\). \[ S_1 = \frac{x^2((x^2)^n - 1)}{x^2 - 1} = \frac{x^2(x^{2n} - 1)}{x^2 - 1} \]
• Series 2: This is a G.P. with first term \(a = \frac{1}{x^2}\) and common ratio \(r = \frac{1}{x^2}\). \[ S_2 = \frac{\frac{1}{x^2}\left(1 - \left(\frac{1}{x^2}\right)^n\right)}{1 - \frac{1}{x^2}} = \frac{\frac{1}{x^2}\left(1 - \frac{1}{x^{2n}}\right)}{\frac{x^2 - 1}{x^2}} = \frac{1 - \frac{1}{x^{2n}}}{x^2 - 1} = \frac{x^{2n} - 1}{x^{2n}(x^2 - 1)} \]
• Series 3: Summing the constant value 2 exactly \(n\) times: \[ S_3 = 2n \]

Step 3: Combining and factoring the expressions.
Adding our three sub-series sums together: \[ S = \frac{x^2(x^{2n} - 1)}{x^2 - 1} + \frac{x^{2n} - 1}{x^{2n}(x^2 - 1)} + 2n \] Factoring out the shared expression \(\frac{x^{2n} - 1}{x^2 - 1}\) from the first two terms: \[ S = \frac{x^{2n} - 1}{x^2 - 1} \left[ x^2 + \frac{1}{x^{2n}} \right] + 2n \] Taking a common denominator inside the square brackets: \[ S = \frac{x^{2n} - 1}{x^2 - 1} \left[ \frac{x^2 \cdot x^{2n} + 1}{x^{2n}} \right] + 2n = \frac{x^{2n} - 1}{x^2 - 1} \times \frac{x^{2n+2} + 1}{x^{2n}} + 2n \] This directly matches the algebraic layout of Option (1).