Find the 16th term in the expansion of `(sqrt(x)-sqrt(y))^(17)`

Answer

To find the 16th term in the expansion of \((\sqrt{x} - \sqrt{y})^{17}\), we can use the Binomial Theorem, which states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, \(a = \sqrt{x}\), \(b = -\sqrt{y}\), and \(n = 17\).

### Step-by-Step Solution:

1. **Identify the term to find**: We need to find the 16th term in the expansion. In terms of \(r\), the 16th term corresponds to \(r = 15\) (since we start counting from \(r = 0\)).

2. **Write the general term**: The general term \(T_{r+1}\) in the expansion can be expressed as:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Substituting \(n = 17\), \(a = \sqrt{x}\), and \(b = -\sqrt{y}\):
\[ T_{r+1} = \binom{17}{r} (\sqrt{x})^{17-r} (-\sqrt{y})^r \]

3. **Substitute \(r = 15\)**: To find the 16th term, substitute \(r = 15\):
\[ T_{16} = \binom{17}{15} (\sqrt{x})^{17-15} (-\sqrt{y})^{15} \]
This simplifies to:
\[ T_{16} = \binom{17}{15} (\sqrt{x})^{2} (-\sqrt{y})^{15} \]

4. **Calculate the binomial coefficient**: The binomial coefficient \(\binom{17}{15}\) is equal to \(\binom{17}{2}\) (since \(\binom{n}{r} = \binom{n}{n-r}\)):
\[ \binom{17}{2} = \frac{17 \times 16}{2 \times 1} = 136 \]

5. **Substitute back into the term**: Now we substitute this back into our expression for \(T_{16}\):
\[ T_{16} = 136 (\sqrt{x})^{2} (-\sqrt{y})^{15} \]

6. **Simplify the expression**: We know that \((\sqrt{x})^{2} = x\) and \((-\sqrt{y})^{15} = -y^{15/2}\):
\[ T_{16} = 136 x (-y^{15/2}) = -136 x y^{15/2} \]

### Final Result:
Thus, the 16th term in the expansion of \((\sqrt{x} - \sqrt{y})^{17}\) is:
\[ \boxed{-136 x y^{15/2}} \]