In the expansion of `(x+(2)/(x^(2)))^(15)` , the term independent of x is

Options

A. `""^(15)C_(2).2^(6)`
B. `""^(15)C_(5).2^(5)`
C. `""^(15)C_(4).2^(4)`
D. None of these

Answer

To find the term independent of \( x \) in the expansion of \( \left( x + \frac{2}{x^2} \right)^{15} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = x \), \( b = \frac{2}{x^2} \), and \( n = 15 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{15}{r} x^{15-r} \left( \frac{2}{x^2} \right)^r \]

2. **Simplify the General Term**:
We can simplify the general term:
\[ T_{r+1} = \binom{15}{r} x^{15-r} \cdot \frac{2^r}{x^{2r}} = \binom{15}{r} \cdot 2^r \cdot x^{15 - r - 2r} = \binom{15}{r} \cdot 2^r \cdot x^{15 - 3r} \]

3. **Find the Term Independent of \( x \)**:
To find the term that is independent of \( x \), we set the exponent of \( x \) to zero:
\[ 15 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 15 \implies r = 5 \]

4. **Substitute \( r \) Back into the General Term**:
Now, we substitute \( r = 5 \) back into the general term to find the term independent of \( x \):
\[ T_{6} = \binom{15}{5} \cdot 2^5 \cdot x^{15 - 3 \cdot 5} \]
Since \( x^{15 - 15} = x^0 \), we have:
\[ T_{6} = \binom{15}{5} \cdot 2^5 \]

5. **Calculate \( \binom{15}{5} \) and \( 2^5 \)**:
We calculate \( \binom{15}{5} \):
\[ \binom{15}{5} = \frac{15!}{5!(15-5)!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003 \]
And \( 2^5 = 32 \).

6. **Final Calculation**:
Now, we multiply these results:
\[ T_{6} = 3003 \cdot 32 = 96096 \]

### Conclusion:
The term independent of \( x \) in the expansion of \( \left( x + \frac{2}{x^2} \right)^{15} \) is \( 96096 \).