If the middle term in the expansion of `(1+x)^(2n)` is the greatest term, then x lies in the interval

Options

A. `n-1 lt x lt n`
B. `(n)/(n+1) lt x lt (n+1)/(n)`
C. `n lt x lt n +1`
D. `(n+1)/(n) lt x lt (n)/(n+1)`

Answer

To solve the problem, we need to find the interval of \( x \) such that the middle term in the expansion of \( (1+x)^{2n} \) is the greatest term.

### Step-by-Step Solution:

1. **Identify the Middle Term**:
The expansion of \( (1+x)^{2n} \) has \( 2n + 1 \) terms. The middle term is given by:
\[ T_{n+1} = \binom{2n}{n} x^n \]

2. **Identify the Greatest Term**:
The greatest term in the binomial expansion occurs when:
\[ \frac{T_k}{T_{k-1}} \leq 1 \]
where \( T_k = \binom{2n}{k} x^k \). The term \( T_k \) is the \( k \)-th term in the expansion.

3. **Set Up the Ratio**:
The ratio of consecutive terms is:
\[ \frac{T_k}{T_{k-1}} = \frac{\binom{2n}{k} x^k}{\binom{2n}{k-1} x^{k-1}} = \frac{2n-k+1}{k} x \]
We want this ratio to be less than or equal to 1 for the greatest term condition.

4. **Apply the Condition**:
For the middle term \( T_{n+1} \) to be the greatest term, we need:
\[ \frac{T_{n+1}}{T_n} \leq 1 \]
This gives us:
\[ \frac{2n - n + 1}{n} x \leq 1 \implies \frac{n + 1}{n} x \leq 1 \]

5. **Solve the Inequality**:
Rearranging the inequality:
\[ x \leq \frac{n}{n + 1} \]

6. **Consider the Other Side**:
For \( T_n \) to be less than or equal to \( T_{n+1} \):
\[ \frac{T_n}{T_{n-1}} \geq 1 \]
This gives us:
\[ \frac{2n - n}{n - 1} x \geq 1 \implies \frac{n}{n - 1} x \geq 1 \]
Rearranging gives:
\[ x \geq \frac{n - 1}{n} \]

7. **Combine the Results**:
From the inequalities, we have:
\[ \frac{n - 1}{n} \leq x \leq \frac{n}{n + 1} \]

8. **Final Interval**:
Therefore, the interval for \( x \) is:
\[ \left( \frac{n-1}{n}, \frac{n}{n+1} \right) \]

### Conclusion:
The interval for \( x \) such that the middle term in the expansion of \( (1+x)^{2n} \) is the greatest term is:
\[ \left( \frac{n-1}{n}, \frac{n}{n+1} \right) \]