1. Problem 1: 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 2: Given $f(x) = \frac{1}{(1+2x)^{1/3}}$, find the binomial expansion up to $x^3$. 3. Problem 3: Approximate $(0.98)^{10}$ using binomial expansion up to $x^4$ and round to 4 decimal places. 4. Problem 4: Given $f(x) = \frac{15}{\sqrt{1-x}}$, find the first 5 terms in the expansion. --- ### Step-by-step solutions: ### 1. Problem 1 1. Write $f(x)$ as $f(x) = (1+2x)^2 \cdot \frac{1}{1-x^2}$. 2. Expand $(1+2x)^2 = 1 + 4x + 4x^2$. 3. Use geometric series for $\frac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n} = 1 + x^2 + x^4 + \cdots$ valid for $|x|<1$. 4. Multiply the two series up to terms of degree 3: $$ (1 + 4x + 4x^2)(1 + x^2) = 1 + 4x + 4x^2 + x^2 + 4x^3 + \text{higher terms} $$ 5. Combine like terms: $$ 1 + 4x + 5x^2 + 4x^3 $$ 6. So, first 4 terms are $1 + 4x + 5x^2 + 4x^3$. 7. The expansion is valid for $|x| < 1$ because of the geometric series. --- ### 2. Problem 2 1. Given $f(x) = (1+2x)^{-1/3}$. 2. Use binomial expansion for $(1 + u)^n = 1 + n u + \frac{n(n-1)}{2} u^2 + \frac{n(n-1)(n-2)}{6} u^3 + \cdots$ where $u=2x$, $n = -\frac{1}{3}$. 3. Calculate coefficients: - $n = -\frac{1}{3}$ - $n(n-1) = -\frac{1}{3} \times (-\frac{4}{3}) = \frac{4}{9}$ - $n(n-1)(n-2) = -\frac{1}{3} \times (-\frac{4}{3}) \times (-\frac{7}{3}) = -\frac{28}{27}$ 4. Substitute: $$ f(x) = 1 - \frac{1}{3} (2x) + \frac{4}{9} \frac{(2x)^2}{2} - \frac{28}{27} \frac{(2x)^3}{6} + \cdots $$ 5. Simplify terms: - $-\frac{1}{3} \times 2x = -\frac{2}{3} x$ - $\frac{4}{9} \times \frac{4x^2}{2} = \frac{4}{9} \times 2x^2 = \frac{8}{9} x^2$ - $-\frac{28}{27} \times \frac{8x^3}{6} = -\frac{28}{27} \times \frac{4}{3} x^3 = -\frac{112}{81} x^3$ 6. Final expansion up to $x^3$: $$ 1 - \frac{2}{3} x + \frac{8}{9} x^2 - \frac{112}{81} x^3 $$ --- ### 3. Problem 3 1. Write $(0.98)^{10} = (1 - 0.02)^{10}$. 2. Use binomial expansion: $$ (1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k $$ with $x = -0.02$, $n=10$. 3. Calculate terms up to $x^4$: - $\binom{10}{0} (-0.02)^0 = 1$ - $\binom{10}{1} (-0.02)^1 = 10 \times (-0.02) = -0.2$ - $\binom{10}{2} (-0.02)^2 = 45 \times 0.0004 = 0.018$ - $\binom{10}{3} (-0.02)^3 = 120 \times (-0.000008) = -0.00096$ - $\binom{10}{4} (-0.02)^4 = 210 \times 0.00000016 = 0.0000336$ 4. Sum terms: $$ 1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8170736 $$ 5. Rounded to 4 decimal places: $0.8171$. --- ### 4. Problem 4 1. Given $f(x) = \frac{15}{\sqrt{1-x}} = 15 (1-x)^{-1/2}$. 2. Use binomial expansion for $(1-x)^n$ with $n = -\frac{1}{2}$: $$ (1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2} (-x)^2 + \frac{n(n-1)(n-2)}{6} (-x)^3 + \frac{n(n-1)(n-2)(n-3)}{24} (-x)^4 + \cdots $$ 3. Calculate coefficients: - $n = -\frac{1}{2}$ - $n(n-1) = -\frac{1}{2} \times (-\frac{3}{2}) = \frac{3}{4}$ - $n(n-1)(n-2) = \frac{3}{4} \times (-\frac{5}{2}) = -\frac{15}{8}$ - $n(n-1)(n-2)(n-3) = -\frac{15}{8} \times (-\frac{7}{2}) = \frac{105}{16}$ 4. Substitute and simplify terms: $$ f(x) = 15 \left[ 1 + \left(-\frac{1}{2}\right)(-x) + \frac{3}{4} \frac{x^2}{2} + \left(-\frac{15}{8}\right) \frac{(-x)^3}{6} + \frac{105}{16} \frac{x^4}{24} + \cdots \right] $$ 5. Simplify each term: - $1$ - $+ \frac{1}{2} x$ - $+ \frac{3}{8} x^2$ - $+ \frac{15}{48} x^3 = \frac{5}{16} x^3$ - $+ \frac{105}{384} x^4 = \frac{35}{128} x^4$ 6. Multiply all terms by 15: $$ 15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4 $$ 7. So, first 5 terms are: $$ 15 + 7.5 x + 5.625 x^2 + 4.6875 x^3 + 4.1015625 x^4 $$