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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Expansion**:
The expression we are working with is \((x + \frac{1}{x})^{2n}\).

2. **Determine the Total Number of Terms**:
The total number of terms in the expansion of \((a + b)^m\) is \(m + 1\). Here, \(m = 2n\), so the total number of terms is \(2n + 1\).

3. **Find the Middle Term**:
Since \(2n\) is even, the middle term will be the \((n + 1)\)th term in the expansion.

4. **Use the Binomial Theorem**:
The general term in the expansion of \((x + \frac{1}{x})^{2n}\) is given by:
\[ T_k = \binom{2n}{k} x^{2n-k} \left(\frac{1}{x}\right)^k = \binom{2n}{k} x^{2n - 2k} \]
where \(k\) is the term number starting from 0.

5. **Substitute for the Middle Term**:
For the middle term, we need to substitute \(k = n\) (since the \((n + 1)\)th term corresponds to \(k = n\)):
\[ T_{n+1} = \binom{2n}{n} x^{2n - 2n} = \binom{2n}{n} x^0 = \binom{2n}{n} \]

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

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

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