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 \( k = 0, 1, 2, 3 \):
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For \( k = 0 \): \[ \binom{10}{0} (2)^{10} \left(\frac{3}{2}x\right)^0 = 1 \cdot 1024 \cdot 1 = 1024 \]
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For \( k = 1 \): \[ \binom{10}{1} (2)^{9} \left(\frac{3}{2}x\right)^{1} = 10 \cdot 512 \cdot \frac{3}{2}x = 10 \cdot 512 \cdot 1.5x = 7680x \]
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For \( k = 2 \): \[ \binom{10}{2} (2)^{8} \left(\frac{3}{2}x\right)^{2} = 45 \cdot 256 \cdot \left(\frac{9}{4}x^2\right) = 45 \cdot 256 \cdot 2.25x^2 = 256 \cdot 101.25x^2 = 25875x^2 \]
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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:
\[ \boxed{1024 + 7680x + 28800x^2 + 63360x^3} \]