Question

Set up the characteristic equation, which is given by r2 − 5r − 6 = 0. The roots of this equation are given by r1 = 6 and r2 = −1. Thus a general solution for the recurrence is given by an = a6 n + b(−1)n . To find a and b, we use the initial conditions. a0 = 1 implies a + b = 1, while a1 = 3 implies 6a − b = 3.

Answer 1

Answer:

Step-by-step explanation:

From your characteristic equation, your recursive equation is

$f(n+2) = 5 f(n=1) +6f(n)$

the general solution:

$f(n) = A6^n + B(-1)^n$

The initial conditions are

$f(0) =1 ~~~and~~~ f(1) = 3$

For f(0) = 1, that is

$f(0) = A6^0 + B(-1)^0 = A+ B=1 (*)$

For f(1) = 3, that is

$f(1) = A6^1 + B(-1)^1 = 3$

$6A- B = 3 (**)$

From (*) and (**) you solve for A and B

you have A = 4/7 and B= 3/7

Replace A, B into the general one, you have the particular solution for the given condition

$f(n) = 4/7 *6^n +3/7*(-1)^n$