Step-by-step Solution:

To find the ratio of the coefficients, we first need to identify the specific coefficient in each binomial expansion using the Binomial Theorem.

According to the Binomial Theorem, the general term in the expansion of \((1+x)^m\) is given by \(T_{k+1} = C(m, k)x^k\), where \(C(m, k)\) is the binomial coefficient, often written as \(\binom{m}{k}\).

1. Find the coefficient of \(x^n\) in the expansion of \((1+x)^{2n}\):
For this expansion, the total power is \(m = 2n\), and we are interested in the term where the power of \(x\) is \(n\), so \(k=n\).
The coefficient is \(C(m, k) = C(2n, n)\).
\[C(2n, n) = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}\]
2. Find the coefficient of \(x^n\) in the expansion of \((1+x)^{2n-1}\):
For this expansion, the total power is \(m = 2n-1\), and we are again interested in the term where the power of \(x\) is \(n\), so \(k=n\).
The coefficient is \(C(m, k) = C(2n-1, n)\).
\[C(2n-1, n) = \frac{(2n-1)!}{n!((2n-1)-n)!} = \frac{(2n-1)!}{n!(n-1)!}\]
3. Calculate the ratio of the two coefficients:
We need to find the ratio \(C(2n, n) : C(2n-1, n)\), which can be calculated by dividing the first coefficient by the second.
\[\text{Ratio} = \frac{C(2n, n)}{C(2n-1, n)} = \frac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}} = \frac{(2n)!}{n!n!} \times \frac{n!(n-1)!}{(2n-1)!}\]
We can cancel one of the \(n!\) terms from the numerator and denominator:
\[\text{Ratio} = \frac{(2n)!}{n!} \times \frac{(n-1)!}{(2n-1)!}\]
Next, we use the property \(m! = m \cdot (m-1)!\) to expand the terms \((2n)!\) and \(n!\):
\[\text{Ratio} = \frac{2n \cdot (2n-1)!}{n \cdot (n-1)!} \times \frac{(n-1)!}{(2n-1)!}\]
Now we can cancel the common factorial terms, \((2n-1)!\) and \((n-1)!\), from the numerator and denominator:
\[\text{Ratio} = \frac{2n}{n} = 2\]
Result:
The ratio of the first coefficient to the second coefficient is 2. This is expressed as the ratio 2:1.