1. **State the problem:** We are given the function $$f(x) = \frac{3x^2 + 16}{(1 - 3x)(2 + x)^2}$$ and the partial fraction decomposition form: $$f(x) = \frac{A}{1 - 3x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}$$ with the condition $$|x| < \frac{1}{3}$$. We need to find the values of constants $$A$$, $$B$$, and $$C$$, show that $$B=0$$, and then find the series expansion of $$f(x)$$ up to the $$x^3$$ term. 2. **Set up the equation for partial fractions:** Multiply both sides by the denominator $$ (1 - 3x)(2 + x)^2 $$ to clear fractions: $$3x^2 + 16 = A(2 + x)^2 + B(1 - 3x)(2 + x) + C(1 - 3x)$$ 3. **Expand terms:** - Expand $$ (2 + x)^2 = 4 + 4x + x^2 $$ - Expand $$ (1 - 3x)(2 + x) = 2 + x - 6x - 3x^2 = 2 - 5x - 3x^2 $$ So the equation becomes: $$3x^2 + 16 = A(4 + 4x + x^2) + B(2 - 5x - 3x^2) + C(1 - 3x)$$ 4. **Collect like terms:** Group coefficients of $$x^2$$, $$x$$, and constants: $$3x^2 + 16 = (A x^2 - 3B x^2) + (4A x - 5B x - 3C x) + (4A + 2B + C)$$ Rewrite: $$3x^2 + 16 = (A - 3B) x^2 + (4A - 5B - 3C) x + (4A + 2B + C)$$ 5. **Equate coefficients:** From the equation, coefficients on both sides must be equal: - For $$x^2$$: $$3 = A - 3B$$ - For $$x$$: $$0 = 4A - 5B - 3C$$ (since no $$x$$ term on left side) - For constant term: $$16 = 4A + 2B + C$$ 6. **Solve the system:** From the first equation: $$A = 3 + 3B$$ Substitute $$A$$ into the second and third equations: - Second: $$0 = 4(3 + 3B) - 5B - 3C = 12 + 12B - 5B - 3C = 12 + 7B - 3C$$ - Third: $$16 = 4(3 + 3B) + 2B + C = 12 + 12B + 2B + C = 12 + 14B + C$$ Rewrite: - $$12 + 7B - 3C = 0 \implies 7B - 3C = -12$$ - $$12 + 14B + C = 16 \implies 14B + C = 4$$ 7. **Express $$C$$ from second equation:** $$C = 4 - 14B$$ Substitute into first: $$7B - 3(4 - 14B) = -12$$ $$7B - 12 + 42B = -12$$ $$49B - 12 = -12$$ $$49B = 0 \implies B = 0$$ 8. **Find $$A$$ and $$C$$:** - $$A = 3 + 3(0) = 3$$ - $$C = 4 - 14(0) = 4$$ Thus, $$A=3$$, $$B=0$$, and $$C=4$$. 9. **Rewrite partial fractions:** $$f(x) = \frac{3}{1 - 3x} + \frac{4}{(2 + x)^2}$$ 10. **Find series expansion up to $$x^3$$:** - Expand $$\frac{3}{1 - 3x}$$ as geometric series: $$\frac{3}{1 - 3x} = 3 \sum_{n=0}^\infty (3x)^n = 3 \left(1 + 3x + 9x^2 + 27x^3 + \cdots \right) = 3 + 9x + 27x^2 + 81x^3 + \cdots$$ - Expand $$\frac{4}{(2 + x)^2}$$: Rewrite denominator: $$(2 + x)^2 = 4 \left(1 + \frac{x}{2} \right)^2$$ So: $$\frac{4}{(2 + x)^2} = \frac{4}{4 (1 + \frac{x}{2})^2} = (1 + \frac{x}{2})^{-2}$$ Use binomial series for negative powers: $$(1 + u)^{-2} = 1 - 2u + 3u^2 - 4u^3 + \cdots$$ with $$u = \frac{x}{2}$$: $$= 1 - 2 \cdot \frac{x}{2} + 3 \left(\frac{x}{2}\right)^2 - 4 \left(\frac{x}{2}\right)^3 + \cdots = 1 - x + \frac{3x^2}{4} - \frac{x^3}{2} + \cdots$$ 11. **Add the two expansions:** $$f(x) = (3 + 9x + 27x^2 + 81x^3) + (1 - x + \frac{3x^2}{4} - \frac{x^3}{2}) + \cdots$$ Combine like terms: - Constant: $$3 + 1 = 4$$ - $$x$$: $$9x - x = 8x$$ - $$x^2$$: $$27x^2 + \frac{3x^2}{4} = \frac{108x^2}{4} + \frac{3x^2}{4} = \frac{111x^2}{4}$$ - $$x^3$$: $$81x^3 - \frac{x^3}{2} = \frac{162x^3}{2} - \frac{x^3}{2} = \frac{161x^3}{2}$$ 12. **Final series expansion up to $$x^3$$:** $$f(x) = 4 + 8x + \frac{111}{4} x^2 + \frac{161}{2} x^3 + \cdots$$ **Summary:** - $$A=3$$, $$B=0$$, $$C=4$$ - Series expansion: $$f(x) = 4 + 8x + \frac{111}{4} x^2 + \frac{161}{2} x^3 + \cdots$$