1. **Problem statement:** Find the first three terms in ascending powers of $x$ for the expansions: (i) $\left(1 + \frac{x}{2}\right)^5$ (ii) $(3 - 2x)^5$ Then use these to find the first three terms of (iii) $(3 - \frac{x}{2} - x^2)^5$. 2. **Formula and rules:** Use the binomial expansion formula: $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 3. **Part (i):** Expand $\left(1 + \frac{x}{2}\right)^5$ up to $x^2$ term. $$ \binom{5}{0} 1^5 \left(\frac{x}{2}\right)^0 = 1 $$ $$ \binom{5}{1} 1^4 \left(\frac{x}{2}\right)^1 = 5 \times \frac{x}{2} = \frac{5x}{2} $$ $$ \binom{5}{2} 1^3 \left(\frac{x}{2}\right)^2 = 10 \times \frac{x^2}{4} = \frac{10x^2}{4} = \frac{5x^2}{2} $$ So the first three terms are: $$ 1 + \frac{5x}{2} + \frac{5x^2}{2} $$ 4. **Part (ii):** Expand $(3 - 2x)^5$ up to $x^2$ term. $$ \binom{5}{0} 3^5 (-2x)^0 = 243 $$ $$ \binom{5}{1} 3^4 (-2x)^1 = 5 \times 81 \times (-2x) = -810x $$ $$ \binom{5}{2} 3^3 (-2x)^2 = 10 \times 27 \times 4x^2 = 1080x^2 $$ So the first three terms are: $$ 243 - 810x + 1080x^2 $$ 5. **Part (iii):** Use the substitution $3 - \frac{x}{2} - x^2 = 3 + b + c$ where $b = -\frac{x}{2}$ and $c = -x^2$. Expand $(3 + b + c)^5$ using the multinomial expansion but only up to $x^2$ terms. The first three terms correspond to: - The term with $b^0 c^0$: $3^5 = 243$ - The term with $b^1 c^0$: $5 \times 3^4 \times b = 5 \times 81 \times \left(-\frac{x}{2}\right) = -\frac{405x}{2}$ - The terms with $b^2 c^0$ and $b^0 c^1$: - $\binom{5}{2} 3^3 b^2 = 10 \times 27 \times \left(-\frac{x}{2}\right)^2 = 10 \times 27 \times \frac{x^2}{4} = \frac{270x^2}{4} = \frac{135x^2}{2}$ - $5 \times 3^4 c = 5 \times 81 \times (-x^2) = -405x^2$ Sum these $x^2$ terms: $$ \frac{135x^2}{2} - 405x^2 = \frac{135x^2}{2} - \frac{810x^2}{2} = -\frac{675x^2}{2} $$ 6. **Final first three terms of (iii):** $$ 243 - \frac{405x}{2} - \frac{675x^2}{2} $$