Consider the expansion of `(sqrt(x)-(2)/(x^(2)))^(16)`

Options

A. There is a term independent of x
B. Coefficinet of `x^(3)=480`
C. There is no term independent of x
D. Coefficient of `x^(3)=240`

Explanation

To find the coefficient of a specific term in the expansion of \((\sqrt{x} - \frac{2}{x^2})^{16}\), we can use the binomial theorem. Let's break down the solution step-by-step.

### Step 1: Identify the general term in the binomial expansion
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = \sqrt{x}\), \(b = -\frac{2}{x^2}\), and \(n = 16\). Therefore, the general term becomes:
\[ T_r = \binom{16}{r} (\sqrt{x})^{16-r} \left(-\frac{2}{x^2}\right)^r \]

### Step 2: Simplify the general term
Now we simplify \(T_r\):
\[ T_r = \binom{16}{r} (\sqrt{x})^{16-r} \cdot (-2)^r \cdot \frac{1}{(x^2)^r} \]
\[ = \binom{16}{r} (-2)^r x^{\frac{16-r}{2} - 2r} \]
\[ = \binom{16}{r} (-2)^r x^{\frac{16 - r - 4r}{2}} = \binom{16}{r} (-2)^r x^{\frac{16 - 5r}{2}} \]

### Step 3: Find the term where the exponent of \(x\) is zero
To find the specific term we are interested in, we set the exponent of \(x\) to zero:
\[ \frac{16 - 5r}{2} = 0 \]
Multiplying both sides by 2:
\[ 16 - 5r = 0 \]
Solving for \(r\):
\[ 5r = 16 \quad \Rightarrow \quad r = \frac{16}{5} \]
Since \(r\) must be a non-negative integer, we need to find the closest integer value for \(r\).

### Step 4: Find the nearest integer value for \(r\)
Since \(\frac{16}{5} = 3.2\), we check the integer values \(r = 3\) and \(r = 4\).

### Step 5: Calculate the coefficients for \(r = 3\) and \(r = 4\)
1. For \(r = 3\):
\[ T_3 = \binom{16}{3} (-2)^3 x^{\frac{16 - 15}{2}} = \binom{16}{3} (-8) x^{\frac{1}{2}} \]
This term does not contribute to the constant term.

2. For \(r = 4\):
\[ T_4 = \binom{16}{4} (-2)^4 x^{\frac{16 - 20}{2}} = \binom{16}{4} (16) x^{-2} \]
This term also does not contribute to the constant term.

### Step 6: Find the correct \(r\) value
We need to find \(r\) such that \(16 - 5r = 0\) gives us a valid integer. The only valid integer solution is \(r = 14\).

### Step 7: Calculate the coefficient for \(r = 14\)
Now we calculate \(T_{14}\):
\[ T_{14} = \binom{16}{14} (-2)^{14} x^{\frac{16 - 70}{2}} = \binom{16}{14} (16^7) x^{-27} \]
This term also does not contribute to the constant term.

### Conclusion
The coefficient of the term where \(x^0\) appears is found using \(r = 14\):
\[ \text{Coefficient} = \binom{16}{14} (-2)^{14} \]
Calculating this gives us the final answer.