Question

recurrence relations: solve the recurrence relation $a_n = 5a_{n - 1}-4a_{n - 2}$ given initial conditions $a_0 = 1, a_1 = 0$.

Explanation:

Step1: Write characteristic equation

$r^2 - 5r + 4 = 0$

Step2: Solve quadratic equation

Factor: $(r-1)(r-4)=0$, so $r_1=1, r_2=4$

Step3: Write general solution

$a_n = C_1(1)^n + C_2(4)^n = C_1 + C_24^n$

Step4: Use $a_0=1$ to find $C_1$

$1 = C_1 + C_24^0 \implies C_1 + C_2 = 1$

Step5: Use $a_1=0$ to find $C_2$

$0 = C_1 + C_24^1 \implies C_1 + 4C_2 = 0$

Step6: Solve for $C_1,C_2$

Subtract Step4 from Step5: $3C_2 = -1 \implies C_2 = -\frac{1}{3}$; then $C_1 = 1 - (-\frac{1}{3}) = \frac{4}{3}$

Step7: Substitute into general solution

$a_n = \frac{4}{3} - \frac{1}{3}4^n$

Answer:

$a_n = \frac{4}{3} - \frac{4^n}{3}$