1. **State the problem:** We want to solve the recurrence relation $$a_n - 2a_{n-1} - 3a_{n-2} = 0$$ for $$n \geq 2$$ with initial conditions $$a_0 = 3$$ and $$a_1 = 1$$ using generating functions. 2. **Define the generating function:** Let $$A(x) = \sum_{n=0}^\infty a_n x^n$$ be the generating function for the sequence $a_n$. 3. **Write the recurrence in terms of generating functions:** Multiply the recurrence by $$x^n$$ and sum over $$n \geq 2$$: $$\sum_{n=2}^\infty a_n x^n - 2 \sum_{n=2}^\infty a_{n-1} x^n - 3 \sum_{n=2}^\infty a_{n-2} x^n = 0$$ 4. **Rewrite sums to express in terms of $$A(x)$$:** - $$\sum_{n=2}^\infty a_n x^n = A(x) - a_0 - a_1 x$$ - $$\sum_{n=2}^\infty a_{n-1} x^n = x \sum_{n=1}^\infty a_n x^{n} = x (A(x) - a_0)$$ - $$\sum_{n=2}^\infty a_{n-2} x^n = x^2 \sum_{n=0}^\infty a_n x^n = x^2 A(x)$$ 5. **Substitute back:** $$A(x) - a_0 - a_1 x - 2x (A(x) - a_0) - 3x^2 A(x) = 0$$ 6. **Group terms:** $$A(x) - 3 - x - 2x A(x) + 6x - 3x^2 A(x) = 0$$ 7. **Collect $$A(x)$$ terms and constants:** $$A(x)(1 - 2x - 3x^2) = 3 + x - 6x = 3 - 5x$$ 8. **Solve for $$A(x)$$:** $$A(x) = \frac{3 - 5x}{1 - 2x - 3x^2}$$ 9. **Factor denominator:** $$1 - 2x - 3x^2 = (1 - 3x)(1 + x)$$ 10. **Partial fraction decomposition:** Assume $$\frac{3 - 5x}{(1 - 3x)(1 + x)} = \frac{A}{1 - 3x} + \frac{B}{1 + x}$$ Multiply both sides by denominator: $$3 - 5x = A(1 + x) + B(1 - 3x) = A + A x + B - 3 B x = (A + B) + (A - 3B) x$$ Equate coefficients: - Constant: $$3 = A + B$$ - Coefficient of $$x$$: $$-5 = A - 3B$$ Solve system: From first, $$A = 3 - B$$ Substitute into second: $$-5 = (3 - B) - 3B = 3 - 4B \Rightarrow -8 = -4B \Rightarrow B = 2$$ Then $$A = 3 - 2 = 1$$ 11. **Rewrite generating function:** $$A(x) = \frac{1}{1 - 3x} + \frac{2}{1 + x}$$ 12. **Use geometric series expansion:** Recall $$\frac{1}{1 - r x} = \sum_{n=0}^\infty r^n x^n$$ for $$|x| < \frac{1}{|r|}$$. So, $$A(x) = \sum_{n=0}^\infty 3^n x^n + 2 \sum_{n=0}^\infty (-1)^n x^n = \sum_{n=0}^\infty (3^n + 2(-1)^n) x^n$$ 13. **Identify the sequence:** $$a_n = 3^n + 2(-1)^n$$ 14. **Check initial conditions:** - $$a_0 = 3^0 + 2(-1)^0 = 1 + 2 = 3$$ - $$a_1 = 3^1 + 2(-1)^1 = 3 - 2 = 1$$ Matches given initial values. **Final answer:** $$\boxed{a_n = 3^n + 2(-1)^n}$$