The `14^(th)` term from the end in the expansion of `(sqrt(x ) - sqrt(y))^(17)` , is

Options

A. `""^(17)C_(5) x^(6) (- sqrt(y))^(5)`
B. `""^(17)C_(6) (sqrt(x))^(11) y^(3)`
C. `""^(17)C_(4) x^(13//2) y^(2)`
D. none of these

Answer

To find the 14th term from the end in the expansion of \((\sqrt{x} - \sqrt{y})^{17}\), we can follow these steps:

### Step 1: Identify the parameters
In the expression \((\sqrt{x} - \sqrt{y})^{17}\), we have:
- \(n = 17\)
- \(a = \sqrt{x}\)
- \(b = -\sqrt{y}\)

### Step 2: Use the formula for the r-th term from the end
The r-th term from the end in a binomial expansion can be found using the formula:
\[ T_{r} = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
To find the 14th term from the end, we need to calculate:
\[ T_{14} = T_{n - (14 - 1)} = T_{17 - 13} = T_{5} \]

### Step 3: Calculate the 5th term from the beginning
Using the formula for the r-th term from the beginning:
\[ T_{r} = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
For the 5th term:
- \(r = 5\)
- \(n = 17\)
- \(a = \sqrt{x}\)
- \(b = -\sqrt{y}\)

Thus, we have:
\[ T_{5} = \binom{17}{5-1} (\sqrt{x})^{17-(5-1)} (-\sqrt{y})^{5-1} \]
This simplifies to:
\[ T_{5} = \binom{17}{4} (\sqrt{x})^{14} (-\sqrt{y})^{4} \]

### Step 4: Simplify the expression
Now, we can simplify:
\[ T_{5} = \binom{17}{4} (\sqrt{x})^{14} (\sqrt{y})^{4} (-1)^{4} \]
Since \((-1)^{4} = 1\), we have:
\[ T_{5} = \binom{17}{4} (\sqrt{x})^{14} (y^{2}) \]

### Step 5: Calculate the binomial coefficient
Now, we need to calculate \(\binom{17}{4}\):
\[ \binom{17}{4} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380 \]

### Step 6: Final expression for the 14th term
Putting it all together, we get:
\[ T_{14} = 2380 \cdot x^{7} \cdot y^{2} \]

### Final Answer
The 14th term from the end in the expansion of \((\sqrt{x} - \sqrt{y})^{17}\) is:
\[ \boxed{2380 x^{7} y^{2}} \]