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

Options

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

Explanation

To find the 15th term in the expansion of \((\sqrt{x} - \sqrt{y})^{17}\), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, we can rewrite the expression as:

\[ (\sqrt{x} - \sqrt{y})^{17} = (\sqrt{x} + (-\sqrt{y}))^{17} \]

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

### Step 1: Identify the general term
The general term \(T_{r+1}\) in the expansion is given by:

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

Substituting the values of \(a\), \(b\), and \(n\):

\[ T_{r+1} = \binom{17}{r} (\sqrt{x})^{17-r} (-\sqrt{y})^r \]

### Step 2: Simplify the general term
This can be simplified further:

\[ T_{r+1} = \binom{17}{r} x^{(17-r)/2} (-1)^r y^{r/2} \]

### Step 3: Find the 15th term
To find the 15th term, we need to set \(r = 14\) (since \(T_{r+1}\) corresponds to \(T_{15}\)):

\[ T_{15} = \binom{17}{14} x^{(17-14)/2} (-1)^{14} y^{14/2} \]

### Step 4: Calculate the binomial coefficient
Calculating \(\binom{17}{14}\):

\[ \binom{17}{14} = \binom{17}{3} = \frac{17!}{3! \cdot 14!} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680 \]

### Step 5: Substitute values into the term
Now substituting back into the term:

\[ T_{15} = 680 \cdot x^{(3)/2} \cdot 1 \cdot y^{7} \]

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

\[ T_{15} = 680 x^{3/2} y^{7} \]