Find the 15th term in the expansion of `"("sqrt(x)-sqrt(y)")"^(17)`

Explanation

To find the 15th term in the expansion of \((\sqrt{x} - \sqrt{y})^{17}\), we can use the Binomial Theorem. The general term in the expansion of \((a + b)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

In our case, we have \(a = \sqrt{x}\), \(b = -\sqrt{y}\), and \(n = 17\). We want to find the 15th term, which corresponds to \(r = 14\) (since the term number \(T_{r+1}\) corresponds to \(r\)).

### Step-by-Step Solution:

1. **Identify the values**:
- \(n = 17\)
- \(r = 14\)
- \(a = \sqrt{x}\)
- \(b = -\sqrt{y}\)

2. **Use the Binomial Theorem formula**:
\[ T_{15} = \binom{17}{14} (\sqrt{x})^{17-14} (-\sqrt{y})^{14} \]

3. **Calculate the binomial coefficient**:
\[ \binom{17}{14} = \binom{17}{3} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680 \]

4. **Calculate the powers**:
- \((\sqrt{x})^{17-14} = (\sqrt{x})^3 = x^{3/2}\)
- \((- \sqrt{y})^{14} = (-1)^{14} (\sqrt{y})^{14} = 1 \cdot y^{7} = y^{7}\)

5. **Combine the results**:
\[ T_{15} = 680 \cdot x^{3/2} \cdot y^{7} \]

6. **Final answer**:
\[ T_{15} = 680 x^{3/2} y^{7} \]