1. **State the problem:** We want to find an explicit formula for the sequence $a_n$ defined by the recurrence relation $$a_n - 8a_{n-1} + 15a_{n-2} = n5^n, \quad n \geq 2$$ with initial conditions $a_0 = 4$ and $a_1 = 12$. 2. **Identify the type of recurrence:** This is a non-homogeneous linear recurrence relation with constant coefficients. 3. **Solve the homogeneous part:** The associated homogeneous recurrence is $$a_n - 8a_{n-1} + 15a_{n-2} = 0.$$ The characteristic equation is $$r^2 - 8r + 15 = 0.$$ Factoring, $$(r - 3)(r - 5) = 0,$$ so the roots are $r=3$ and $r=5$. 4. **General solution of homogeneous:** $$a_n^h = A3^n + B5^n,$$ where $A$ and $B$ are constants to be determined. 5. **Find a particular solution:** Since the non-homogeneous term is $n5^n$, and $5$ is a root of the characteristic equation, we try a particular solution of the form $$a_n^p = n(Cn + D)5^n = (Cn^2 + Dn)5^n,$$ where $C$ and $D$ are constants to be found. 6. **Substitute $a_n^p$ into the recurrence:** Calculate $a_n^p$, $a_{n-1}^p$, and $a_{n-2}^p$: $$a_n^p = (Cn^2 + Dn)5^n,$$ $$a_{n-1}^p = (C(n-1)^2 + D(n-1))5^{n-1},$$ $$a_{n-2}^p = (C(n-2)^2 + D(n-2))5^{n-2}.$$ Substitute into the left side: $$a_n^p - 8a_{n-1}^p + 15a_{n-2}^p = n5^n.$$ Divide both sides by $5^{n-2}$ to simplify: $$5^2(Cn^2 + Dn) - 8 imes 5 (C(n-1)^2 + D(n-1)) + 15 (C(n-2)^2 + D(n-2)) = n5^2,$$ which simplifies to $$25(Cn^2 + Dn) - 40 (C(n-1)^2 + D(n-1)) + 15 (C(n-2)^2 + D(n-2)) = 25 n.$$ 7. **Expand and simplify:** Expand each term: - $25(Cn^2 + Dn) = 25Cn^2 + 25Dn$ - $-40(C(n-1)^2 + D(n-1)) = -40C(n^2 - 2n + 1) - 40D(n - 1) = -40Cn^2 + 80Cn - 40C - 40Dn + 40D$ - $15(C(n-2)^2 + D(n-2)) = 15C(n^2 - 4n + 4) + 15D(n - 2) = 15Cn^2 - 60Cn + 60C + 15Dn - 30D$ Sum all: $$25Cn^2 + 25Dn - 40Cn^2 + 80Cn - 40C - 40Dn + 40D + 15Cn^2 - 60Cn + 60C + 15Dn - 30D = 25n.$$ Group like terms: - $n^2$ terms: $25C n^2 - 40C n^2 + 15C n^2 = 0$ - $n$ terms: $25D n - 40D n + 15D n + 80C n - 60C n = (0) n + (20C) n = (20C + 0) n$ - Constants: $-40C + 40D + 60C - 30D = (20C + 10D)$ So the left side simplifies to $$ (20C) n + (20C + 10D) = 25 n.$$ 8. **Equate coefficients:** For the equation to hold for all $n$, - Coefficient of $n$: $20C = 25 \\ \Rightarrow C = \frac{25}{20} = \frac{5}{4}$ - Constant term: $20C + 10D = 0 \\ \Rightarrow 20 \times \frac{5}{4} + 10D = 0 \\ 25 + 10D = 0 \\ 10D = -25 \\ D = -\frac{25}{10} = -\frac{5}{2}$ 9. **Particular solution:** $$a_n^p = \left( \frac{5}{4} n^2 - \frac{5}{2} n \right) 5^n.$$ 10. **General solution:** $$a_n = a_n^h + a_n^p = A3^n + B5^n + \left( \frac{5}{4} n^2 - \frac{5}{2} n \right) 5^n.$$ 11. **Use initial conditions to solve for $A$ and $B$:** - For $n=0$: $$a_0 = A3^0 + B5^0 + \left( \frac{5}{4} 0^2 - \frac{5}{2} 0 \right) 5^0 = A + B = 4.$$ - For $n=1$: $$a_1 = A3^1 + B5^1 + \left( \frac{5}{4} 1^2 - \frac{5}{2} 1 \right) 5^1 = 3A + 5B + \left( \frac{5}{4} - \frac{5}{2} \right) 5 = 3A + 5B + \left( -\frac{5}{4} \right) 5 = 3A + 5B - \frac{25}{4} = 12.$$ Rewrite second equation: $$3A + 5B = 12 + \frac{25}{4} = \frac{48}{4} + \frac{25}{4} = \frac{73}{4}.$$ From first equation: $B = 4 - A$. Substitute into second: $$3A + 5(4 - A) = \frac{73}{4} \\ 3A + 20 - 5A = \frac{73}{4} \\ -2A = \frac{73}{4} - 20 = \frac{73}{4} - \frac{80}{4} = -\frac{7}{4} \\ A = \frac{7}{8}.$$ Then $$B = 4 - \frac{7}{8} = \frac{32}{8} - \frac{7}{8} = \frac{25}{8}.$$ 12. **Final explicit formula:** $$\boxed{a_n = \frac{7}{8} 3^n + \frac{25}{8} 5^n + \left( \frac{5}{4} n^2 - \frac{5}{2} n \right) 5^n}.$$