15th term in the expansion of `(sqrt(2)-sqrt(y))^(17)` is :

Options

A. `87040y^(7)`
B. `-87040y^(3//2)`
C. `680y^(7)`
D. `-860y^(7//2)`

Explanation

To find the 15th term in the expansion of \((\sqrt{2} - \sqrt{y})^{17}\), we can use the Binomial Theorem. According to the theorem, the \(r + 1\)th term in the expansion of \((x + a)^n\) is given by:

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

### Step-by-step Solution:

1. **Identify the parameters**:
Here, \(n = 17\), \(x = \sqrt{2}\), and \(a = -\sqrt{y}\). We are looking for the 15th term, which means \(r = 14\) (since \(r + 1 = 15\)).

2. **Write the formula for the 15th term**:
\[ T_{15} = \binom{17}{14} (\sqrt{2})^{17-14} (-\sqrt{y})^{14} \]

3. **Calculate the binomial coefficient**:
\[ \binom{17}{14} = \binom{17}{3} = \frac{17!}{3!(17-3)!} = \frac{17!}{3! \cdot 14!} \]
Simplifying this:
\[ \binom{17}{3} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = \frac{4080}{6} = 680 \]

4. **Calculate the powers**:
\[ (\sqrt{2})^{3} = 2^{3/2} = 2^{1.5} = 2 \cdot \sqrt{2} \]
\[ (-\sqrt{y})^{14} = (-1)^{14} (\sqrt{y})^{14} = y^{7} \]

5. **Combine all parts**:
Now substituting back into the term:
\[ T_{15} = 680 \cdot (2 \cdot \sqrt{2}) \cdot y^{7} \]
\[ = 680 \cdot 2 \cdot \sqrt{2} \cdot y^{7} = 1360 \sqrt{2} y^{7} \]

Thus, the 15th term in the expansion of \((\sqrt{2} - \sqrt{y})^{17}\) is:

\[ \boxed{1360 \sqrt{2} y^{7}} \]