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:
\[ (1 + \frac{1}{x^2})^n (1 + x^2)^n \]
We can rewrite this as:
\[ \left( \frac{x^2 + 1}{x^2} \right)^n (1 + x^2)^n = \frac{(x^2 + 1)^n}{(x^2)^n} (1 + x^2)^n \]
This simplifies to:
\[ \frac{(1 + x^2)^n (1 + x^2)^n}{x^{2n}} = \frac{(1 + x^2)^{2n}}{x^{2n}} \]

### Step 2: Identify the general term
The general term \(T_k\) in the expansion of \((1 + x^2)^{2n}\) can be expressed as:
\[ T_k = \binom{2n}{k} (x^2)^k (1)^{2n-k} = \binom{2n}{k} x^{2k} \]

### Step 3: Determine the middle term
To find the middle term, we need to determine the total number of terms in the expansion. The number of terms in \((1 + x^2)^{2n}\) is \(2n + 1\), which is odd. Therefore, the middle term is the \((n + 1)\)th term.

### Step 4: Calculate the middle term
The middle term \(T_{n}\) (since we start counting from \(T_0\)) is given by:
\[ T_n = \binom{2n}{n} x^{2n} \]
Now, substituting this back into our expression, we have:
\[ T_n = \frac{\binom{2n}{n} x^{2n}}{x^{2n}} = \binom{2n}{n} \]

### Final Answer
Thus, the middle term in the expansion is:
\[ \binom{2n}{n} \]