What is the ratio of coefficient of `x^(15)` to the term independent of x in `( x^(2) + ( 2)/( x))^(15)` ?

Options

A. `1//64`
B. `1//32`
C. `1//16`
D. `(1)/(4)`

Answer

To solve the problem, we need to find the ratio of the coefficient of \(x^{15}\) to the term independent of \(x\) in the expression \((x^2 + \frac{2}{x})^{15}\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \(T_{r+1}\) in the binomial expansion of \((x^2 + \frac{2}{x})^{15}\) can be expressed as:
\[ T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{2}{x}\right)^r \]
Simplifying this, we get:
\[ T_{r+1} = \binom{15}{r} x^{2(15-r)} \cdot \frac{2^r}{x^r} = \binom{15}{r} 2^r x^{30 - 3r} \]

2. **Find the Coefficient of \(x^{15}\)**:
To find the coefficient of \(x^{15}\), we set the exponent of \(x\) equal to 15:
\[ 30 - 3r = 15 \]
Solving for \(r\):
\[ 3r = 30 - 15 \implies 3r = 15 \implies r = 5 \]
Now substituting \(r = 5\) back into the general term:
\[ T_{6} = \binom{15}{5} 2^5 x^{15} \]
The coefficient of \(x^{15}\) is:
\[ \binom{15}{5} \cdot 2^5 \]

3. **Find the Term Independent of \(x\)**:
The term independent of \(x\) occurs when the exponent of \(x\) is zero:
\[ 30 - 3r = 0 \]
Solving for \(r\):
\[ 3r = 30 \implies r = 10 \]
Now substituting \(r = 10\) back into the general term:
\[ T_{11} = \binom{15}{10} 2^{10} x^{0} \]
The term independent of \(x\) is:
\[ \binom{15}{10} \cdot 2^{10} \]

4. **Calculate the Ratio**:
Now we need to find the ratio of the coefficient of \(x^{15}\) to the term independent of \(x\):
\[ \text{Ratio} = \frac{\binom{15}{5} \cdot 2^5}{\binom{15}{10} \cdot 2^{10}} \]
Since \(\binom{15}{5} = \binom{15}{10}\), the ratio simplifies to:
\[ \text{Ratio} = \frac{2^5}{2^{10}} = \frac{1}{2^{10-5}} = \frac{1}{2^5} = \frac{1}{32} \]

### Final Answer:
The ratio of the coefficient of \(x^{15}\) to the term independent of \(x\) is \(\frac{1}{32}\).