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

Options

A. 1
B. 2
C. `(1)/(2)`
D. `(1)/(n)`

Explanation

To find the ratio \( \frac{A}{B} \) 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 follow these steps:

### Step 1: Identify the Coefficients
The coefficient of \( x^n \) in the expansion of \( (1 + x)^{2n} \) is given by the binomial coefficient:
\[ A = \binom{2n}{n} \]
Similarly, the coefficient of \( x^n \) in the expansion of \( (1 + x)^{2n-1} \) is:
\[ B = \binom{2n-1}{n} \]

### Step 2: Write the Ratio
Now, we need to find the ratio \( \frac{A}{B} \):
\[ \frac{A}{B} = \frac{\binom{2n}{n}}{\binom{2n-1}{n}} \]

### Step 3: Substitute the Binomial Coefficient Formula
Using the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
We can express \( A \) and \( B \) as:
\[ A = \frac{(2n)!}{n! (2n - n)!} = \frac{(2n)!}{n! n!} \]
\[ B = \frac{(2n-1)!}{n! ((2n-1) - n)!} = \frac{(2n-1)!}{n! (n-1)!} \]

### Step 4: Substitute into the Ratio
Now substituting \( A \) and \( B \) into the ratio:
\[ \frac{A}{B} = \frac{\frac{(2n)!}{n! n!}}{\frac{(2n-1)!}{n! (n-1)!}} = \frac{(2n)! \cdot (n-1)!}{(2n-1)! \cdot n!} \]

### Step 5: Simplify the Ratio
We can simplify this further:
\[ \frac{A}{B} = \frac{(2n) \cdot (2n-1)! \cdot (n-1)!}{(2n-1)! \cdot n \cdot (n-1)!} = \frac{2n}{n} = 2 \]

### Conclusion
Thus, we find that:
\[ \frac{A}{B} = 2 \]