1. Problem Q1: Given $f(x) = \frac{(1 + 2x)^2}{(1 - x)^2}$ (i) Find the first 4 terms in the power series expansion. (ii) State when the expansion of $f(x)$ is valid. 2. Problem Q2: Given $f(x) = \frac{1}{(1 + 2x)^{1/3}}$, find the binomial expansion up to $x^3$. 3. Problem Q3: Approximate $(0.98)^{10}$ using binomial expansion up to $x^4$ and round to 4 decimal places. 4. Problem Q4: Given $f(x) = \frac{15}{\sqrt{1 - x}}$, find the first 5 terms in the expansion. --- ### Q1(i) Expansion of $f(x) = \frac{(1 + 2x)^2}{(1 - x)^2}$ 1. Write $f(x)$ as $f(x) = (1 + 2x)^2 (1 - x)^{-2}$. 2. Expand $(1 + 2x)^2$ using binomial theorem: $$ (1 + 2x)^2 = 1 + 4x + 4x^2 $$ 3. Expand $(1 - x)^{-2}$ using binomial series: $$ (1 - x)^{-2} = \sum_{n=0}^\infty \binom{n+1}{1} x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots $$ 4. Multiply the two series up to $x^3$ terms: $$ f(x) \approx (1 + 4x + 4x^2)(1 + 2x + 3x^2 + 4x^3) $$ Calculate term-by-term: - Constant: $1 \times 1 = 1$ - $x$: $1 \times 2x + 4x \times 1 = 2x + 4x = 6x$ - $x^2$: $1 \times 3x^2 + 4x \times 2x + 4x^2 \times 1 = 3x^2 + 8x^2 + 4x^2 = 15x^2$ - $x^3$: $1 \times 4x^3 + 4x \times 3x^2 + 4x^2 \times 2x = 4x^3 + 12x^3 + 8x^3 = 24x^3$ So first 4 terms: $$ f(x) \approx 1 + 6x + 15x^2 + 24x^3 $$ --- ### Q1(ii) Validity of expansion The binomial series for $(1 - x)^{-2}$ converges for $|x| < 1$. Since $(1 + 2x)^2$ is a polynomial, the radius of convergence is determined by $(1 - x)^{-2}$. Therefore, the expansion is valid for: $$ |x| < 1 $$ --- ### Q2 Binomial expansion of $f(x) = (1 + 2x)^{-1/3}$ up to $x^3$ 1. Use binomial series for $(1 + u)^n$ where $n = -\frac{1}{3}$ and $u = 2x$: $$ (1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \cdots $$ 2. Calculate coefficients: - $n = -\frac{1}{3}$ - $n(n-1) = -\frac{1}{3} \times (-\frac{4}{3}) = \frac{4}{9}$ - $n(n-1)(n-2) = -\frac{1}{3} \times (-\frac{4}{3}) \times (-\frac{7}{3}) = -\frac{28}{27}$ 3. Substitute and simplify: $$ f(x) \approx 1 - \frac{1}{3} (2x) + \frac{4}{9} \frac{(2x)^2}{2} - \frac{28}{27} \frac{(2x)^3}{6} $$ Calculate each term: - $-\frac{1}{3} \times 2x = -\frac{2}{3} x$ - $\frac{4}{9} \times \frac{4x^2}{2} = \frac{4}{9} \times 2x^2 = \frac{8}{9} x^2$ - $-\frac{28}{27} \times \frac{8x^3}{6} = -\frac{28}{27} \times \frac{4}{3} x^3 = -\frac{112}{81} x^3$ 4. Final expansion: $$ f(x) \approx 1 - \frac{2}{3} x + \frac{8}{9} x^2 - \frac{112}{81} x^3 $$ --- ### Q3 Approximate $(0.98)^{10}$ using binomial expansion 1. Write $0.98 = 1 - 0.02$, so: $$ (0.98)^{10} = (1 - 0.02)^{10} $$ 2. Use binomial expansion up to $x^4$ where $x = 0.02$: $$ (1 - x)^{10} = \sum_{k=0}^4 \binom{10}{k} (-x)^k $$ 3. Calculate terms: - $k=0$: $1$ - $k=1$: $10 \times (-0.02) = -0.2$ - $k=2$: $\binom{10}{2} ( -0.02)^2 = 45 \times 0.0004 = 0.018$ - $k=3$: $\binom{10}{3} (-0.02)^3 = 120 \times (-0.000008) = -0.00096$ - $k=4$: $\binom{10}{4} ( -0.02)^4 = 210 \times 0.00000016 = 0.0000336$ 4. Sum terms: $$ 1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8170736 $$ 5. Rounded to 4 decimal places: $$ 0.8171 $$ --- ### Q4 Expansion of $f(x) = \frac{15}{\sqrt{1 - x}}$ 1. Recall binomial expansion for $(1 - x)^{-1/2}$: $$ (1 - x)^{-1/2} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} (-x)^n $$ 2. First 5 terms coefficients: - $n=0$: $1$ - $n=1$: $\frac{1}{2} x$ - $n=2$: $\frac{1 \times 3}{2 \times 4} x^2 = \frac{3}{8} x^2$ - $n=3$: $\frac{1 \times 3 \times 5}{2 \times 4 \times 6} x^3 = \frac{5}{16} x^3$ - $n=4$: $\frac{1 \times 3 \times 5 \times 7}{2 \times 4 \times 6 \times 8} x^4 = \frac{35}{128} x^4$ 3. Multiply by 15: $$ f(x) = 15 \left(1 + \frac{1}{2} x + \frac{3}{8} x^2 + \frac{5}{16} x^3 + \frac{35}{128} x^4 \right) $$ 4. Simplify: $$ f(x) = 15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4 $$ --- Final answers: Q1(i): $1 + 6x + 15x^2 + 24x^3$ Q1(ii): Valid for $|x| < 1$ Q2: $1 - \frac{2}{3} x + \frac{8}{9} x^2 - \frac{112}{81} x^3$ Q3: Approximate $(0.98)^{10} \approx 0.8171$ Q4: $15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4$