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

Options

A. `((n)!)/((n/2)^2!`
B. `((2n)!)/((n/2)^2!`
C. `(1.3.5. . . (2n+1))/(n!)*2^n`
D. `((2n)!)/(n!`

Answer

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

### Step 1: Simplify the Expression
The expression can be rewritten as:
\[ (x^2 + 2 + \frac{1}{x^2})^n \]
Notice that \(x^2 + 2 + \frac{1}{x^2}\) can be expressed as \((x + \frac{1}{x})^2\) since:
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]
Thus, we have:
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]
So, the expression becomes:
\[ ((x + \frac{1}{x})^2)^n = (x + \frac{1}{x})^{2n} \]

### Step 2: Determine the Number of Terms
The expansion of \((x + \frac{1}{x})^{2n}\) will have \(2n + 1\) terms. This is because the number of terms in the expansion of \((a + b)^m\) is \(m + 1\).

### Step 3: Find the Middle Term
Since there are \(2n + 1\) terms, the middle term will be the \((n + 1)\)-th term.

### Step 4: Write the General Term
The general term \(T_r\) in the expansion of \((x + \frac{1}{x})^{2n}\) is given by:
\[ T_r = \binom{2n}{r} (x)^{2n - r} \left(\frac{1}{x}\right)^r = \binom{2n}{r} x^{2n - 2r} \]
where \(r\) varies from \(0\) to \(2n\).

### Step 5: Substitute for the Middle Term
For the middle term, we set \(r = n\):
\[ T_{n+1} = \binom{2n}{n} x^{2n - 2n} = \binom{2n}{n} x^0 = \binom{2n}{n} \]

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