Write down the first three terms is the following expansions
`(1 + 4x)^(-4)`

Answer

To find the first three terms of the expansion of \((1 + 4x)^{-4}\), we can use the Binomial Theorem for negative exponents. The Binomial Theorem states that:

\[ (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \]

For negative exponents, we can express it as:

\[ (1 + x)^{-p} = \sum_{k=0}^{\infty} \binom{-p}{k} x^k \]

Where \(\binom{-p}{k} = (-1)^k \binom{p+k-1}{k}\).

### Step 1: Identify the values
In our case, we have:
- \(x = 4x\)
- \(p = 4\)

### Step 2: Write the first three terms
We will compute the first three terms of the expansion:

1. **First term (k = 0)**:
\[ \binom{-4}{0} (4x)^0 = 1 \]

2. **Second term (k = 1)**:
\[ \binom{-4}{1} (4x)^1 = -4 \cdot 4x = -16x \]

3. **Third term (k = 2)**:
\[ \binom{-4}{2} (4x)^2 = \frac{-4 \cdot (-5)}{2!} \cdot (16x^2) = \frac{20}{2} \cdot 16x^2 = 10 \cdot 16x^2 = 160x^2 \]

### Step 3: Combine the terms
Now, we can combine these terms to write the final expression:

\[ (1 + 4x)^{-4} \approx 1 - 16x + 160x^2 \]

### Final Answer
Thus, the first three terms of the expansion of \((1 + 4x)^{-4}\) are:

\[ 1 - 16x + 160x^2 \]

---