If for positive integers `r gt 1, n gt 2`, the coefficients of the `(3r)`th and `(r+2)`th powers of `x` in the expansion of `(1+x)^(2n)` are equal, then prove that `n=2r+1`.

Answer

To solve the problem, we need to show that if the coefficients of the \(3r\)th and \((r + 2)\)th powers of \(x\) in the expansion of \((1 + x)^{2n}\) are equal, then \(n = 2r + 1\).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficient of \(x^{k}\) in the expansion of \((1 + x)^{2n}\) is given by \(\binom{2n}{k}\). Therefore, we need to find:
- The coefficient of \(x^{3r}\): \(\binom{2n}{3r}\)
- The coefficient of \(x^{r + 2}\): \(\binom{2n}{r + 2}\)

We are given that these coefficients are equal:
\[ \binom{2n}{3r} = \binom{2n}{r + 2} \]

2. **Using the Property of Binomial Coefficients**:
The property of binomial coefficients states that \(\binom{n}{k} = \binom{n}{n-k}\). Thus, we can rewrite the equation:
\[ \binom{2n}{3r} = \binom{2n}{2n - 3r} \quad \text{and} \quad \binom{2n}{r + 2} = \binom{2n}{2n - (r + 2)} \]
This gives us:
\[ 2n - 3r = r + 2 \]

3. **Rearranging the Equation**:
Now, rearranging the equation \(2n - 3r = r + 2\):
\[ 2n = 4r + 2 \]
Dividing both sides by 2:
\[ n = 2r + 1 \]

4. **Conclusion**:
We have shown that \(n = 2r + 1\), which is what we needed to prove.