In the expansion of `(3-sqrt(17/4+3sqrt2))^15` the 11th term is a

Options

A. an irrational number
B. a rational number
C. a positive integer
D. a negative integer

Explanation

To find the 11th term in the expansion of \( (3 - \sqrt{17/4 + 3\sqrt{2}})^{15} \), we can use the Binomial Theorem. The \( n \)-th term in the expansion of \( (a + b)^n \) is given by:

\[ T_k = \binom{n}{k} a^{n-k} b^k \]

In our case, we have:

- \( n = 15 \)
- \( a = 3 \)
- \( b = -\sqrt{17/4 + 3\sqrt{2}} \)
- We are looking for the 11th term, which corresponds to \( k = 10 \) (since the first term corresponds to \( k = 0 \)).

### Step 1: Identify the 11th term
The 11th term \( T_{11} \) can be expressed as:

\[ T_{11} = \binom{15}{10} (3)^{15-10} \left(-\sqrt{17/4 + 3\sqrt{2}}\right)^{10} \]

### Step 2: Calculate the binomial coefficient
Now, we calculate the binomial coefficient:

\[ \binom{15}{10} = \binom{15}{5} = \frac{15!}{10!5!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003 \]

### Step 3: Calculate \( (3)^{15-10} \)
Next, calculate \( (3)^{15-10} \):

\[ (3)^{5} = 243 \]

### Step 4: Calculate \( \left(-\sqrt{17/4 + 3\sqrt{2}}\right)^{10} \)
Next, we need to compute \( \left(-\sqrt{17/4 + 3\sqrt{2}}\right)^{10} \):

Since \( (-\sqrt{x})^{10} = (\sqrt{x})^{10} = x^5 \), we need to find \( \left(\sqrt{17/4 + 3\sqrt{2}}\right)^{10} \):

\[ \left(\sqrt{17/4 + 3\sqrt{2}}\right)^{10} = \left(17/4 + 3\sqrt{2}\right)^{5} \]

### Step 5: Combine all parts to find \( T_{11} \)
Now, we can combine all parts to find \( T_{11} \):

\[ T_{11} = 3003 \cdot 243 \cdot \left(17/4 + 3\sqrt{2}\right)^{5} \]

### Step 6: Analyze \( \left(17/4 + 3\sqrt{2}\right)^{5} \)
The term \( 17/4 + 3\sqrt{2} \) contains an irrational component \( 3\sqrt{2} \). Therefore, raising it to the 5th power will still yield an irrational number.

### Step 7: Determine the nature of \( T_{11} \)
Since \( T_{11} \) is a product of a rational number \( 3003 \cdot 243 \) and an irrational number \( \left(17/4 + 3\sqrt{2}\right)^{5} \), the result will be irrational.

Thus, the 11th term is an irrational number.

### Final Answer
The 11th term is an irrational number.