If the middle term in the expansion of `(1 + x)^(2n)` is `(1.3.5…(2n - 1))/(n!) k^(n).x^(n)`, the value of k is

Options

A. 4
B. 1
C. 0
D. 2

Answer

To solve the problem, we need to find the value of \( k \) in the expression for the middle term of the expansion of \( (1 + x)^{2n} \).

### Step-by-Step Solution:

1. **Identify the Middle Term**:
The middle term in the expansion of \( (1 + x)^{2n} \) can be expressed as:
\[ T_{n+1} = \binom{2n}{n} x^n \]
where \( T_{n+1} \) is the \( (n+1)^{th} \) term.

2. **Calculate \( \binom{2n}{n} \)**:
The binomial coefficient \( \binom{2n}{n} \) can be calculated using the formula:
\[ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \]

3. **Express \( (2n)! \)**:
The factorial \( (2n)! \) can be expressed in terms of products of odd and even numbers:
\[ (2n)! = 1 \cdot 2 \cdot 3 \cdots (2n) = (1 \cdot 3 \cdots (2n-1)) \cdot (2 \cdot 4 \cdots (2n)) \]
The product of even numbers can be factored out as:
\[ 2^n \cdot (1 \cdot 2 \cdots n) = 2^n \cdot n! \]

4. **Combine the Expressions**:
Thus, we can rewrite \( (2n)! \) as:
\[ (2n)! = (1 \cdot 3 \cdots (2n-1)) \cdot 2^n \cdot n! \]

5. **Substituting Back**:
Now substituting this back into the expression for \( \binom{2n}{n} \):
\[ \binom{2n}{n} = \frac{(1 \cdot 3 \cdots (2n-1)) \cdot 2^n \cdot n!}{(n!)^2} = \frac{1 \cdot 3 \cdots (2n-1) \cdot 2^n}{n!} \]

6. **Final Form of the Middle Term**:
Therefore, the middle term can be expressed as:
\[ T_{n+1} = \frac{1 \cdot 3 \cdots (2n-1) \cdot 2^n}{n!} x^n \]

7. **Comparing with Given Expression**:
The problem states that the middle term is:
\[ \frac{1 \cdot 3 \cdots (2n-1)}{n!} k^n x^n \]
By comparing both expressions, we have:
\[ \frac{1 \cdot 3 \cdots (2n-1) \cdot 2^n}{n!} x^n = \frac{1 \cdot 3 \cdots (2n-1)}{n!} k^n x^n \]

8. **Solve for \( k \)**:
From the comparison, we can deduce:
\[ k^n = 2^n \]
Taking the \( n^{th} \) root on both sides gives:
\[ k = 2 \]

### Conclusion:
Thus, the value of \( k \) is \( \boxed{2} \).