If A and B are coefficients of `x^(n)` in the expansion of `(1+x)^(2n)` and `(1+x)^(2n-1)` respectively, then find the value of `(A)/(B)`.

Answer

To find the value of \( \frac{A}{B} \), where \( A \) and \( B \) are the coefficients of \( x^n \) in the expansions of \( (1+x)^{2n} \) and \( (1+x)^{2n-1} \) respectively, we can use the Binomial Theorem.

### Step-by-step Solution:

1. **Identify the Coefficient \( A \)**:
The coefficient \( A \) of \( x^n \) in the expansion of \( (1+x)^{2n} \) can be found using the Binomial Theorem:
\[ A = \binom{2n}{n} \]
This represents the number of ways to choose \( n \) items from \( 2n \) items.

2. **Identify the Coefficient \( B \)**:
The coefficient \( B \) of \( x^n \) in the expansion of \( (1+x)^{2n-1} \) is given by:
\[ B = \binom{2n-1}{n} \]
This represents the number of ways to choose \( n \) items from \( 2n-1 \) items.

3. **Formulate the Ratio \( \frac{A}{B} \)**:
Now, we need to find the ratio \( \frac{A}{B} \):
\[ \frac{A}{B} = \frac{\binom{2n}{n}}{\binom{2n-1}{n}} \]

4. **Use the Property of Binomial Coefficients**:
We can express the binomial coefficients in terms of factorials:
\[ \binom{2n}{n} = \frac{(2n)!}{n! \cdot n!} \]
\[ \binom{2n-1}{n} = \frac{(2n-1)!}{n! \cdot (n-1)!} \]
Substituting these into the ratio gives:
\[ \frac{A}{B} = \frac{\frac{(2n)!}{n! \cdot n!}}{\frac{(2n-1)!}{n! \cdot (n-1)!}} = \frac{(2n)! \cdot (n-1)!}{(2n-1)! \cdot n!} \]

5. **Simplify the Expression**:
We know that \( (2n)! = (2n)(2n-1)! \), so substituting this in gives:
\[ \frac{A}{B} = \frac{(2n)(2n-1)! \cdot (n-1)!}{(2n-1)! \cdot n!} \]
The \( (2n-1)! \) cancels out:
\[ \frac{A}{B} = \frac{(2n)(n-1)!}{n!} \]
Since \( n! = n \cdot (n-1)! \), we can further simplify:
\[ \frac{A}{B} = \frac{2n}{n} = 2 \]

### Final Answer:
Thus, the value of \( \frac{A}{B} \) is \( 2 \).