prove that the coefficient of `x^n` in the expansion of `(1+x)^(2n)` is twice the coefficient of `x^n` in the expansion of `(1+x)^(2n-1)`

Explanation

To prove that the coefficient of \(x^n\) in the expansion of \((1+x)^{2n}\) is twice the coefficient of \(x^n\) in the expansion of \((1+x)^{2n-1}\), we will follow these steps:

### Step 1: Identify the coefficients in the binomial expansions
The general term in the binomial expansion of \((1+x)^k\) is given by:
\[ T_r = \binom{k}{r} x^r \]
Thus, the coefficient of \(x^n\) in \((1+x)^{2n}\) is:
\[ \text{Coefficient of } x^n \text{ in } (1+x)^{2n} = \binom{2n}{n} \]

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

### Step 2: Establish the relationship between the coefficients
We need to show that:
\[ \binom{2n}{n} = 2 \cdot \binom{2n-1}{n} \]

### Step 3: Use the property of binomial coefficients
We can express the binomial coefficient \(\binom{2n}{n}\) in terms of \(\binom{2n-1}{n}\):
\[ \binom{2n}{n} = \frac{(2n)!}{n!n!} \]
And,
\[ \binom{2n-1}{n} = \frac{(2n-1)!}{n!(n-1)!} \]

### Step 4: Relate the two coefficients
Now, we can relate \(\binom{2n}{n}\) to \(\binom{2n-1}{n}\):
\[ \binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2n)(2n-1)!}{n!n!} = \frac{2n}{n} \cdot \frac{(2n-1)!}{(n-1)!n!} \]
This simplifies to:
\[ \binom{2n}{n} = 2 \cdot \frac{(2n-1)!}{(n-1)!n!} = 2 \cdot \binom{2n-1}{n} \]

### Conclusion
Thus, we have shown that:
\[ \binom{2n}{n} = 2 \cdot \binom{2n-1}{n} \]
This proves that the coefficient of \(x^n\) in the expansion of \((1+x)^{2n}\) is indeed twice the coefficient of \(x^n\) in the expansion of \((1+x)^{2n-1}\).