Answer:
f) a[n] = -(-2)^n +2^n
g) a[n] = (1/2)((-2)^-n +2^-n)
Explanation:
Both of these problems are solved in the same way. The characteristic equation comes from ...
a[n] -k²·a[n-2] = 0
Using a[n] = r^n, we have ...
r^n -k²r^(n-2) = 0
r^(n-2)(r² -k²) = 0
r² -k² = 0
r = ±k
a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q
We find p and q from the initial conditions.
__
f) k² = 4, so k = 2.
a[0] = 0 = p + q
a[1] = 4 = -2p +2q
Dividing the second equation by 2 and adding the first, we have ...
2 = 2q
q = 1
p = -1
The solution is a[n] = -(-2)^n +2^n.
__
g) k² = 1/4, so k = 1/2.
a[0] = 1 = p + q
a[1] = 0 = -p/2 +q/2
Multiplying the first equation by 1/2 and adding the second, we get ...
1/2 = q
p = 1 -q = 1/2
Using k = 2^-1, we can write the solution as follows.
The solution is a[n] = (1/2)((-2)^-n +2^-n).