If the middle term in the expreansion of `(1+x)^(2n)` is `([1.3.5….(2n-1)])/(n!) (k)` , where n is a positive integer , then `k` is .

Answer

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

### Step 1: Identify the Middle Term
In the expansion of \( (1+x)^{2n} \), since \( 2n \) is even, the middle term is given by the formula for the \( \left( n + 1 \right) \)-th term.

### Step 2: Write the General Term
The general term \( T_r \) in the binomial expansion of \( (1+x)^{2n} \) is given by:
\[ T_r = \binom{2n}{r} x^r \]
where \( r \) is the term number.

### Step 3: Find the Middle Term
For \( n = 2n \), the middle term corresponds to \( r = n \). Therefore, the middle term \( T_{n+1} \) is:
\[ T_{n+1} = \binom{2n}{n} x^n \]

### Step 4: Calculate \( \binom{2n}{n} \)
Using the formula for binomial coefficients:
\[ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \]

### Step 5: Express the Middle Term
Substituting this into our expression for the middle term, we get:
\[ T_{n+1} = \frac{(2n)!}{(n!)^2} x^n \]

### Step 6: Compare with Given Expression
We are given that the middle term can also be expressed as:
\[ \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \cdot k \]

### Step 7: Express the Product of Odd Numbers
The product of the first \( n \) odd numbers can be expressed as:
\[ 1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{(2n)!}{2^n n!} \]
This is a known result.

### Step 8: Set Up the Equation
Now we can equate the two expressions for the middle term:
\[ \frac{(2n)!}{(n!)^2} x^n = \frac{(2n)!}{2^n n!} \cdot k \]

### Step 9: Simplify the Equation
Cancel \( (2n)! \) from both sides:
\[ \frac{1}{(n!)^2} x^n = \frac{1}{2^n n!} \cdot k \]

### Step 10: Solve for \( k \)
Multiplying both sides by \( 2^n n! \):
\[ k = \frac{2^n x^n}{n!} \]

### Conclusion
Thus, the value of \( k \) is:
\[ k = 2^n x^n \]