Answer:
The solution to the recurrence relation is given by an = 2^(n+1) - 1.
Explanation:
(a) To solve the recurrence relation an = an-2 + 4, with initial conditions a1 = 3 and a2 = 5, we'll consider two cases: one for when n is even and one for when n is odd.
For n even:
Substituting n = 2k (where k is a positive integer) into the recurrence relation, we get:
a2k = a2k-2 + 4
Now let's write out a few terms to observe the pattern:
a2 = a0 + 4
a4 = a2 + 4
a6 = a4 + 4
...
We notice that a2k = a0 + 4k for even values of k.
Using the initial condition a2 = 5, we can find a0:
a2 = a0 + 4(1)
5 = a0 + 4
a0 = 1
Therefore, for even values of n, the solution is given by an = 1 + 4k.
For n odd:
Substituting n = 2k + 1 (where k is a non-negative integer) into the recurrence relation, we get:
a2k+1 = a2k-1 + 4
Again, let's write out a few terms to observe the pattern:
a3 = a1 + 4
a5 = a3 + 4
a7 = a5 + 4
...
We see that a2k+1 = a1 + 4k for odd values of k.
Using the initial condition a1 = 3, we find:
a3 = a1 + 4(1)
a3 = 3 + 4
a3 = 7
Therefore, for odd values of n, the solution is given by an = 3 + 4k.
(b) To solve the recurrence relation an = 2an-1 + 1, with initial condition a1 = 1, we'll find a general expression for an.
Let's write out a few terms to observe the pattern:
a2 = 2a1 + 1
a3 = 2a2 + 1
a4 = 2a3 + 1
...
We can see that each term is one more than twice the previous term.
By substituting repeatedly, we can express an in terms of a1:
an = 2(2(2(...2(a1) + 1)...)) + 1
= 2^n * a1 + (2^n - 1)
Using the initial condition a1 = 1, we have:
an = 2^n * 1 + (2^n - 1)
= 2^n + 2^n - 1
= 2 * 2^n - 1
Therefore, the solution to the recurrence relation is given by an = 2^(n+1) - 1.