The middle term in the expansion of `(1+(1)/(x^(2)))^(n) ( 1+x^(2))^(n)` is

Options

A. `""^(2n)C_(n)x^(2n)`
B. `""^(2n)C_(n)x^(-2n)`
C. `""^(2n) C_(n)`
D. `""^(2n) C_(n-1)`

Answer

To find the middle term in the expansion of \((1 + \frac{1}{x^2})^n (1 + x^2)^n\), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ (1 + \frac{1}{x^2})^n (1 + x^2)^n \]
This can be rewritten as:
\[ \left(\frac{1}{x^2}\right)^n (1 + x^2)^n = \frac{1}{x^{2n}} (1 + x^2)^n \]

### Step 2: Identify the Total Number of Terms
The expansion of \((1 + x^2)^n\) will have \(n + 1\) terms. Therefore, the total number of terms in the product \(\frac{1}{x^{2n}} (1 + x^2)^n\) will also be \(n + 1\).

### Step 3: Determine the Middle Term
Since the total number of terms \(n + 1\) is odd, the middle term will be the \(\left(\frac{n + 1}{2} + 1\right)\)th term, which simplifies to the \(\left(\frac{n + 1}{2}\right)\)th term.

### Step 4: Find the General Term
The general term \(T_k\) in the expansion of \((1 + x^2)^n\) is given by:
\[ T_k = \binom{n}{k} (x^2)^k = \binom{n}{k} x^{2k} \]
Thus, the middle term \(T_{\frac{n}{2}}\) will be:
\[ T_{\frac{n}{2}} = \binom{n}{\frac{n}{2}} x^{n} \]

### Step 5: Combine with \(\frac{1}{x^{2n}}\)
Now, we multiply this middle term by \(\frac{1}{x^{2n}}\):
\[ \text{Middle Term} = \frac{1}{x^{2n}} \cdot \binom{n}{\frac{n}{2}} x^{n} = \binom{n}{\frac{n}{2}} \frac{x^{n}}{x^{2n}} = \binom{n}{\frac{n}{2}} \frac{1}{x^{n}} \]

### Final Result
Thus, the middle term in the expansion of \((1 + \frac{1}{x^2})^n (1 + x^2)^n\) is:
\[ \frac{\binom{n}{\frac{n}{2}}}{x^n} \]