Find the first three terms in the expansion of `[ 2+ x ( 3+ 4x)]^(5)` in ascending powers of `x`.

Answer

To find the first three terms in the expansion of \((2 + x(3 + 4x))^5\) in ascending powers of \(x\), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case:
- \(a = 2\)
- \(b = x(3 + 4x)\)
- \(n = 5\)

### Step 1: Identify the terms for expansion

We need to find the first three terms of the expansion. According to the Binomial Theorem, the first three terms are given by:

1. \(T_0 = \binom{5}{0} a^5 b^0\)
2. \(T_1 = \binom{5}{1} a^4 b^1\)
3. \(T_2 = \binom{5}{2} a^3 b^2\)

### Step 2: Calculate the first term \(T_0\)

\[ T_0 = \binom{5}{0} \cdot 2^5 \cdot (x(3 + 4x))^0 = 1 \cdot 32 \cdot 1 = 32 \]

### Step 3: Calculate the second term \(T_1\)

\[ T_1 = \binom{5}{1} \cdot 2^4 \cdot (x(3 + 4x))^1 = 5 \cdot 16 \cdot (3x + 4x^2) \]
\[ = 80x + 20x^2 \]

### Step 4: Calculate the third term \(T_2\)

\[ T_2 = \binom{5}{2} \cdot 2^3 \cdot (x(3 + 4x))^2 = 10 \cdot 8 \cdot (3x + 4x^2)^2 \]

Calculating \((3x + 4x^2)^2\):

\[ (3x + 4x^2)^2 = 9x^2 + 24x^3 + 16x^4 \]

Thus,

\[ T_2 = 80 \cdot (9x^2 + 24x^3 + 16x^4) = 720x^2 + 1920x^3 + 1280x^4 \]

### Step 5: Combine the terms

Now we combine the first three terms:

\[ T_0 + T_1 + T_2 = 32 + (80x + 20x^2) + (720x^2 + 1920x^3 + 1280x^4) \]

### Final Result

The first three terms in the expansion of \((2 + x(3 + 4x))^5\) in ascending powers of \(x\) are:

\[ 32 + 80x + 740x^2 \]