Find the middle term in the expansion of `(x^2+1/(x^2)+2)^ndot`

Explanation

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

### Step 1: Rewrite the Expression
We start by rewriting the expression inside the parentheses. We notice that:
\[ x^2 + \frac{1}{x^2} + 2 = (x + \frac{1}{x})^2 \]
This is because:
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]
So, we can express our original expression as:
\[ (x + \frac{1}{x})^2 \]

### Step 2: Raise to the Power of n
Now we raise this expression to the power of \(n\):
\[ (x + \frac{1}{x})^{2n} \]

### Step 3: Determine the Number of Terms
In the expansion of \((x + \frac{1}{x})^{2n}\), there will be \(2n + 1\) terms. This is because the binomial expansion of \((a + b)^m\) has \(m + 1\) terms.

### Step 4: Identify the Middle Term
Since there are \(2n + 1\) terms, the middle term will be the \((n + 1)\)th term.

### Step 5: Use the Binomial Theorem
According to the Binomial Theorem, the \(k\)th term in the expansion of \((a + b)^m\) is given by:
\[ T_k = \binom{m}{k-1} a^{m-(k-1)} b^{k-1} \]
In our case, \(a = x\), \(b = \frac{1}{x}\), and \(m = 2n\). Therefore, the middle term \(T_{n+1}\) is:
\[ T_{n+1} = \binom{2n}{n} x^{2n-n} \left(\frac{1}{x}\right)^{n} = \binom{2n}{n} x^{n} \cdot \frac{1}{x^n} = \binom{2n}{n} \]

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