Step 1 : Understanding the Question:
This problem asks us to calculate the sum of a series up to $n$ terms, where each term is the square of a binomial of the form $\left(x^k + \frac{1}{x^k}\right)$. To find the sum, we can expand each binomial term, group similar algebraic terms together, and identify the resulting geometric progressions. By summing these individual parts, we can find the simplified expression for the total sum of the series.
Step 2 : Key Formulas and Approach:
The expansion of each term in the series uses the algebraic identity:
\[ (A + B)^2 = A^2 + 2AB + B^2 \]
Applying this to the $k$-th term of the series, we get:
\[ \left(x^k + \frac{1}{x^k}\right)^2 = x^{2k} + 2 + \frac{1}{x^{2k}} \]
This allows us to split the sum of $n$ terms into three separate sub-series:
1. A geometric progression with term $x^{2k}$.
2. A geometric progression with term $\frac{1}{x^{2k}}$.
3. A constant term of 2 summed $n$ times.
The sum of a geometric progression with first term $a$ and common ratio $r$ is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
We will evaluate these three sums separately and then combine and simplify them.
Step 3 : Detailed Explanation:
We expand each term of the series individually.
The first term is: $\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$.
The second term is: $\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + 2 + \frac{1}{x^4}$.
The third term is: $\left(x^3 + \frac{1}{x^3}\right)^2 = x^6 + 2 + \frac{1}{x^6}$, and so on.
We write the total sum $S$ of $n$ terms by grouping the expanded terms into three separate series: $S = \sum_{k=1}^n x^{2k} + \sum_{k=1}^n \frac{1}{x^{2k}} + \sum_{k=1}^n 2$.
The first sub-series is a geometric progression: $S_1 = x^2 + x^4 + x^6 + \dots + x^{2n}$. The first term is $a_1 = x^2$ and the common ratio is $r_1 = x^2$. Using the GP sum formula, we get: $S_1 = \frac{x^2(x^{2n} - 1)}{x^2 - 1}$.
The second sub-series is also a geometric progression: $S_2 = \frac{1}{x^2} + \frac{1}{x^4} + \dots + \frac{1}{x^{2n}}$. The first term is $a_2 = \frac{1}{x^2}$ and the common ratio is $r_2 = \frac{1}{x^2}$. Applying the GP sum formula gives: $S_2 = \frac{\frac{1}{x^2}\left(1 - \frac{1}{x^{2n}}\right)}{1 - \frac{1}{x^2}} = \frac{x^{2n} - 1}{x^{2n}(x^2 - 1)}$.
The third sub-series is the constant 2 summed $n$ times: $S_3 = 2n$.
Now, we add the three sums together: $S = \frac{x^2(x^{2n} - 1)}{x^2 - 1} + \frac{x^{2n} - 1}{x^{2n}(x^2 - 1)} + 2n$.
We factor out the common term $\frac{x^{2n} - 1}{x^2 - 1}$ from the first two terms: $S = \frac{x^{2n} - 1}{x^2 - 1} \left[ x^2 + \frac{1}{x^{2n}} \right] + 2n$.
Simplifying the expression inside the brackets by finding a common denominator gives: $S = \frac{x^{2n} - 1}{x^2 - 1} \left[ \frac{x^{2n+2} + 1}{x^{2n}} \right] + 2n$.
Step 4 : Final Answer:
The sum of the series is $\frac{x^{2n} - 1}{x^2 - 1} \times \frac{x^{2n+2} + 1}{x^{2n}} + 2n$, which corresponds to Option (A).