Question

Expand (2+3/2x)^10 in ascending powers of x up to x^3

Answer 1

To expand \( (2 + \frac{3}{2}x)^{10} \) in ascending powers of \( x \) up to \( x^3 \), we can use the binomial theorem, which states that:

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

For our expression, we have \( a = 2 \), \( b = \frac{3}{2}x \), and \( n = 10 \).

### Step 1: Identify terms

We want to expand and only need terms up to \( x^3 \).

### Step 2: Calculate relevant terms

We will calculate the terms for \( k = 0, 1, 2, 3 \):

1. **For \( k = 0 \)**:
\[
\binom{10}{0} (2)^{10} \left(\frac{3}{2}x\right)^0 = 1 \cdot 1024 \cdot 1 = 1024
\]

2. **For \( k = 1 \)**:
\[
\binom{10}{1} (2)^{9} \left(\frac{3}{2}x\right)^{1} = 10 \cdot 512 \cdot \frac{3}{2}x = 10 \cdot 512 \cdot 1.5x = 7680x
\]

3. **For \( k = 2 \)**:
\[
\binom{10}{2} (2)^{8} \left(\frac{3}{2}x\right)^{2} = 45 \cdot 256 \cdot \left(\frac{9}{4}x^2\right) = 45 \cdot 256 \cdot 2.25x^2 = 256 \cdot 101.25x^2 = 25875x^2
\]

4. **For \( k = 3 \)**:
\[
\binom{10}{3} (2)^{7} \left(\frac{3}{2}x\right)^{3} = 120 \cdot 128 \cdot \left(\frac{27}{8}x^3\right) = 120 \cdot 128 \cdot 3.375x^3 = 120 \cdot 128 \cdot 3.375x^3 = 51600x^3
\]

### Step 3: Combine the terms

Now we can sum all the relevant terms together:

\[
(2 + \frac{3}{2}x)^{10} \approx 1024 + 7680x + 28800x^2 + 63360x^3
\]

### Final result

Thus, the expansion of \( (2 + \frac{3}{2}x)^{10} \) in ascending powers of \( x \) up to \( x^3 \) is:

\[
\boxed{1024 + 7680x + 28800x^2 + 63360x^3}
\]