FPM(2023) January Paper 02R Q1 and solution

Problem and Solution

Part (a): Expand $ \left( 1 + \frac{x}{4} \right)^8 $ in ascending powers of $ x $

Using the binomial expansion formula:

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

we expand $ \left( 1 + \frac{x}{4} \right)^8 $ with $ a = 1 $, $ b = \frac{x}{4} $, and $ n = 8 $.

Step 1: Write the terms of the expansion up to $ x^3 $:

$$ \left( 1 + \frac{x}{4} \right)^8 = \binom{8}{0} \cdot 1^8 + \binom{8}{1} \cdot 1^7 \cdot \frac{x}{4} + \binom{8}{2} \cdot 1^6 \cdot \left( \frac{x}{4} \right)^2 + \binom{8}{3} \cdot 1^5 \cdot \left( \frac{x}{4} \right)^3 + \dots $$

Step 2: Compute each term:

• Constant term: $$ \binom{8}{0} \cdot 1^8 = 1 $$
• Term in $ x $: $$ \binom{8}{1} \cdot 1^7 \cdot \frac{x}{4} = 8 \cdot \frac{x}{4} = 2x $$
• Term in $ x^2 $: $$ \binom{8}{2} \cdot 1^6 \cdot \left( \frac{x}{4} \right)^2 = \frac{8 \cdot 7}{2} \cdot \frac{x^2}{16} = 28 \cdot \frac{x^2}{16} = \frac{7x^2}{4} $$
• Term in $ x^3 $: $$ \binom{8}{3} \cdot 1^5 \cdot \left( \frac{x}{4} \right)^3 = \frac{8 \cdot 7 \cdot 6}{6} \cdot \frac{x^3}{64} = 56 \cdot \frac{x^3}{64} = \frac{7x^3}{8} $$

Step 3: Combine the terms:

$$ \left( 1 + \frac{x}{4} \right)^8 = 1 + 2x + \frac{7x^2}{4} + \frac{7x^3}{8} + \dots $$

Part (b): Approximate $ (1.035)^8 $

We know that:

$$ 1.035 = 1 + \frac{x}{4}, \quad \text{so } x = 4 \cdot (1.035 - 1) = 0.14 $$

Step 1: Substitute $ x = 0.14 $ into the expansion:

$$ (1.035)^8 \approx 1 + 2x + \frac{7x^2}{4} + \frac{7x^3}{8} $$

Step 2: Compute each term:

• Constant term: $ 1 $
• Term in $ x $: $$ 2x = 2(0.14) = 0.28 $$
• Term in $ x^2 $: $$ \frac{7x^2}{4} = \frac{7(0.14)^2}{4} = \frac{7(0.0196)}{4} = \frac{0.1372}{4} = 0.0343 $$
• Term in $ x^3 $: $$ \frac{7x^3}{8} = \frac{7(0.14)^3}{8} = \frac{7(0.002744)}{8} = \frac{0.019208}{8} = 0.002401 $$

Step 3: Add the terms:

$$ (1.035)^8 \approx 1 + 0.28 + 0.0343 + 0.002401 = 1.3167 $$

Final Answer

$$ (1.035)^8 \approx \boxed{1.3167} $$