The coefficient of `x^3` in `(1-4x)^(1//2)` is

Options

A. `-4`
B. 2
C. `-8`
D. 12

Answer

To find the coefficient of \( x^3 \) in the expansion of \( (1 - 4x)^{1/2} \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{\infty} \binom{n}{k} a^{n-k} b^k \]

For our case, we can rewrite \( (1 - 4x)^{1/2} \) as:

\[ (1 + (-4x))^{1/2} \]

Here, \( a = 1 \), \( b = -4x \), and \( n = \frac{1}{2} \).

### Step 1: Identify the general term
The general term \( T_k \) in the expansion is given by:

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

Substituting our values, we get:

\[ T_k = \binom{\frac{1}{2}}{k} (1)^{\frac{1}{2}-k} (-4x)^k = \binom{\frac{1}{2}}{k} (-4)^k x^k \]

### Step 2: Find the term for \( k = 3 \)
We need to find the coefficient of \( x^3 \), so we set \( k = 3 \):

\[ T_3 = \binom{\frac{1}{2}}{3} (-4)^3 x^3 \]

### Step 3: Calculate \( \binom{\frac{1}{2}}{3} \)
The binomial coefficient \( \binom{n}{k} \) is given by:

\[ \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \]

For \( n = \frac{1}{2} \) and \( k = 3 \):

\[ \binom{\frac{1}{2}}{3} = \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right)}{3!} = \frac{\frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right)}{6} \]

Calculating this:

\[ = \frac{\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2}}{6} = \frac{\frac{3}{8}}{6} = \frac{3}{48} = \frac{1}{16} \]

### Step 4: Calculate \( T_3 \)
Now substituting back into \( T_3 \):

\[ T_3 = \frac{1}{16} (-4)^3 x^3 = \frac{1}{16} (-64) x^3 = -4 x^3 \]

### Step 5: Coefficient of \( x^3 \)
Thus, the coefficient of \( x^3 \) in the expansion of \( (1 - 4x)^{1/2} \) is:

\[ \text{Coefficient of } x^3 = -4 \]

### Final Answer
The coefficient of \( x^3 \) in \( (1 - 4x)^{1/2} \) is \( \boxed{-4} \).

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