1. **Problem 1:** Given $f(x)=\frac{(1+2x)^2}{1 - x^2}$, find the first 4 terms in the power series expansion and state when the expansion is valid. 2. **Step 1:** Expand numerator and denominator separately. Numerator: $(1+2x)^2 = 1 + 4x + 4x^2$ Denominator: $\frac{1}{1 - x^2} = \sum_{n=0}^\infty x^{2n} = 1 + x^2 + x^4 + \cdots$ 3. **Step 2:** Multiply the series (up to 4 terms in total): $f(x) = (1 + 4x + 4x^2)(1 + x^2 + x^4 + \cdots)$ Up to terms with $x^3$: $= 1 + 4x + 4x^2 + x^2 + 4x^3 + \cdots = 1 + 4x + 5x^2 + 4x^3 + \cdots$ 4. **Step 3:** First 4 terms: $$f(x) \approx 1 + 4x + 5x^2 + 4x^3$$ 5. **Step 4:** Interval of validity comes from denominator $1 - x^2 \neq 0$ and use of geometric series: Power series for $\frac{1}{1 - x^2}$ converges when $|x^2| < 1 \Rightarrow |x| < 1$. Since numerator is polynomial, expansion valid **for $|x| < 1$**. --- 6. **Problem 2:** Given $f(x) = \frac{1}{(1+2x)^{1/3}}$, find the binomial expansion up to $x^3$. 7. **Step 1:** Use generalized binomial theorem: For $f(x) = (1 + u)^k$, expansion: $$ (1+u)^k = 1 + ku + \frac{k(k-1)}{2!}u^2 + \frac{k(k-1)(k-2)}{3!}u^3 + \cdots $$ Here, $k = -\frac{1}{3}$ and $u = 2x$. 8. **Step 2:** Calculate coefficients: - First term: $1$ - Second: $k u = -\frac{1}{3} \times 2x = -\frac{2}{3}x$ - Third: $\frac{k(k-1)}{2} u^2 = \frac{-\frac{1}{3}(-\frac{4}{3})}{2} (2x)^2 = \frac{\frac{4}{9}}{2} \times 4x^2 = \frac{4}{9} \times 2 x^2 = \frac{8}{9} x^2$ - Fourth: $\frac{k(k-1)(k-2)}{6} u^3 = \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} (2x)^3$ Calculate numerator: $-\frac{1}{3} \times -\frac{4}{3} = \frac{4}{9}$ Then multiply by $-\frac{7}{3}$: $\frac{4}{9} \times -\frac{7}{3} = -\frac{28}{27}$ Divide by 6: $-\frac{28}{27} \times \frac{1}{6} = -\frac{28}{162} = -\frac{14}{81}$ Multiply by $ (2x)^3 = 8x^3$: $-\frac{14}{81} \times 8x^3 = -\frac{112}{81} x^3$ 9. **Step 3:** Write expansion up to $x^3$: $$f(x) \approx 1 - \frac{2}{3}x + \frac{8}{9} x^2 - \frac{112}{81} x^3$$ --- 10. **Problem 3:** Given $(0.98)^{10}$, use binomial expansion with $x=-0.02$, expand up to $x^4$, approximate to 4 decimal places. 11. **Step 1:** Write $(0.98)^{10} = (1 - 0.02)^{10}$ Use binomial theorem: $ (1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k$ Here, $n=10$, $x = -0.02$ 12. **Step 2:** Calculate terms up to $x^4$: - $\binom{10}{0} (-0.02)^0 = 1$ - $\binom{10}{1} (-0.02)^1 = 10 \times -0.02 = -0.2$ - $\binom{10}{2} (-0.02)^2 = 45 \times 0.0004 = 0.018$ - $\binom{10}{3} (-0.02)^3 = 120 \times (-0.000008) = -0.00096$ - $\binom{10}{4} (-0.02)^4 = 210 \times 0.00000016 = 0.0000336$ 13. **Step 3:** Sum approximation: $1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8170736$ Rounded to 4 decimal places: $$0.8171$$ --- 14. **Problem 4:** Given $f(x) = \frac{15}{\sqrt{1-x}}$, find the first 5 terms of the expansion. 15. **Step 1:** Write $f(x) = 15(1-x)^{-1/2}$. Use binomial expansion for $(1 - x)^k$ with $k = -\frac{1}{2}$: $$ (1 - x)^k = 1 + k(-x) + \frac{k(k-1)}{2!}(-x)^2 + \frac{k(k-1)(k-2)}{3!}(-x)^3 + \frac{k(k-1)(k-2)(k-3)}{4!}(-x)^4 + \cdots $$ 16. **Step 2:** Calculate terms one by one: - Term 0: $1$ - Term 1: $k(-x) = -\frac{1}{2} \times (-x) = \frac{1}{2}x$ - Term 2: $\frac{k(k-1)}{2} (-x)^2 = \frac{-\frac{1}{2}(-\frac{3}{2})}{2} x^2 = \frac{\frac{3}{4}}{2} x^2 = \frac{3}{8} x^2$ - Term 3: $\frac{k(k-1)(k-2)}{6}(-x)^3 = \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{6}(-x)^3$ Calculate numerator: $-\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4}$ Multiply by $-\frac{5}{2}$: $\frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8}$ Divide by 6: $-\frac{15}{8} \times \frac{1}{6} = -\frac{15}{48} = -\frac{5}{16}$ Multiply by $(-x)^3 = -x^3$: $-\frac{5}{16} \times -x^3 = \frac{5}{16} x^3$ - Term 4: $\frac{k(k-1)(k-2)(k-3)}{24} (-x)^4$ Calculate numerator: $k(k-1)(k-2)(k-3) = -\frac{1}{2} \times -\frac{3}{2} \times -\frac{5}{2} \times -\frac{7}{2}$ Stepwise: $-\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4}$ $\frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8}$ $-\frac{15}{8} \times -\frac{7}{2} = \frac{105}{16}$ Divide by $24$: $\frac{105}{16} \times \frac{1}{24} = \frac{105}{384} = \frac{35}{128}$ Multiply by $(-x)^4 = x^4$: $\frac{35}{128} x^4$ 17. **Step 3:** Combine terms and multiply entire series by 15: $$ f(x) = 15 \left(1 + \frac{1}{2} x + \frac{3}{8} x^2 + \frac{5}{16} x^3 + \frac{35}{128} x^4\right) = 15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4 $$ --- **Summary:** - Q1: First 4 terms $1 + 4x + 5x^2 + 4x^3$, valid for $|x|<1$. - Q2: Binomial expansion $1 - \frac{2}{3}x + \frac{8}{9} x^2 - \frac{112}{81} x^3$. - Q3: Approximate $(0.98)^{10} \approx 0.8171$. - Q4: First 5 terms $15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4$.