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)`.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the coefficients \( A \) and \( B \) from the expansions of \( (1+x)^{2n} \) and \( (1+x)^{2n-1} \) respectively, and then calculate the ratio \( \frac{A}{B} \).

### Step-by-Step Solution:

1. **Identify the Coefficient \( A \)**:
The coefficient of \( x^n \) in the expansion of \( (1+x)^{2n} \) can be found using the binomial theorem:
\[ A = \binom{2n}{n} \]

2. **Identify the Coefficient \( B \)**:
Similarly, the coefficient of \( x^n \) in the expansion of \( (1+x)^{2n-1} \) is:
\[ B = \binom{2n-1}{n} \]

3. **Calculate 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 use the property of binomial coefficients:
\[ \binom{2n}{n} = \frac{(2n)!}{n! \cdot n!} \]
and
\[ \binom{2n-1}{n} = \frac{(2n-1)!}{n! \cdot (2n-1-n)!} = \frac{(2n-1)!}{n! \cdot (n-1)!} \]

5. **Substituting the Values**:
Substitute these values into the ratio:
\[ \frac{A}{B} = \frac{\frac{(2n)!}{n! \cdot n!}}{\frac{(2n-1)!}{n! \cdot (n-1)!}} = \frac{(2n)! \cdot (n-1)!}{(2n-1)! \cdot n!} \]

6. **Simplifying the Ratio**:
Notice that \( (2n)! = (2n)(2n-1)! \):
\[ \frac{A}{B} = \frac{(2n)(2n-1)! \cdot (n-1)!}{(2n-1)! \cdot n!} = \frac{2n \cdot (n-1)!}{n!} \]
Since \( n! = n \cdot (n-1)! \), we can simplify further:
\[ \frac{A}{B} = \frac{2n \cdot (n-1)!}{n \cdot (n-1)!} = \frac{2n}{n} = 2 \]

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