The coefficient of `x^(4)` in `((x)/(2)-(3)/(x^(2)))^(10)` is :

Options

A. `405/256`
B. `504/259`
C. `450/263`
D. None of these

Answer

To find the coefficient of \( x^4 \) in the expression \( \left( \frac{x}{2} - \frac{3}{x^2} \right)^{10} \), we can use the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Identify the general term using the Binomial Theorem
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
In our case, we can set:
- \( a = \frac{x}{2} \)
- \( b = -\frac{3}{x^2} \)
- \( n = 10 \)

The general term \( T_{r+1} \) in the expansion is given by:
\[ T_{r+1} = \binom{10}{r} \left( \frac{x}{2} \right)^{10-r} \left( -\frac{3}{x^2} \right)^r \]

### Step 2: Simplify the general term
Now, simplifying \( T_{r+1} \):
\[ T_{r+1} = \binom{10}{r} \left( \frac{x^{10-r}}{2^{10-r}} \right) \left( -\frac{3^r}{x^{2r}} \right) \]
This can be rewritten as:
\[ T_{r+1} = \binom{10}{r} \left( -3 \right)^r \frac{x^{10-r}}{2^{10-r}} \cdot \frac{1}{x^{2r}} = \binom{10}{r} \left( -3 \right)^r \frac{x^{10 - r - 2r}}{2^{10 - r}} = \binom{10}{r} \left( -3 \right)^r \frac{x^{10 - 3r}}{2^{10 - r}} \]

### Step 3: Find the value of \( r \) for which the power of \( x \) is 4
We need the exponent of \( x \) to be 4:
\[ 10 - 3r = 4 \]
Solving for \( r \):
\[ 10 - 4 = 3r \implies 6 = 3r \implies r = 2 \]

### Step 4: Substitute \( r \) back into the general term
Now, substitute \( r = 2 \) into the general term:
\[ T_{3} = \binom{10}{2} \left( -3 \right)^2 \frac{x^{10 - 3 \cdot 2}}{2^{10 - 2}} = \binom{10}{2} \cdot 9 \cdot \frac{x^4}{2^8} \]

### Step 5: Calculate the coefficient
Now calculate \( \binom{10}{2} \):
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]
Thus, the coefficient becomes:
\[ 45 \cdot 9 \cdot \frac{1}{256} = \frac{405}{256} \]

### Final Answer
The coefficient of \( x^4 \) in the expansion is:
\[ \frac{405}{256} \]