Assume that the given expression is divisible by 4 (i.e., P(n) is true). Then we need to show that implies the expression w/ n -> n + 1 is also divisible by 4 (i.e., show P(n) => P(n + 1)...)

Let's replace n -> (n + 1) in our given expression:

(3 ^ (2(n + 1) + 1)) + 5 can be rewritten as 3 ^ (2n + 3) + 5

==> 3 ^ [(2n + 1) + 2] + 5

BUT: 3 ^ [(2n + 1) + 2] = (3 ^ (2n + 1)) * 3^2, which becomes 9(3 ^ (2n + 1))

Substituting this in, we arrive at:

==> 9(3 ^ (2n + 1)) + 5 for expression. We can break this apart to form

==> 8(3 ^ (2n + 1)) + 3 ^ (2n + 1) + 5 ==> 8(3 ^ (2n + 1)) + [3 ^ (2n + 1) + 5]

However, note that we assumed the expression in brackets (which is P(n)) was divisible by 4 (at the start), and obviously 8(3 ^ (2n + 1)) is divisible by 4, so the whole expression must also be divisible by 4.

In other words P(n) => P(n + 1) for our induction hypothesis.

Now just go and check that P(1) works (i.e., that the expression is divisible by 4 when n = 1), and that should give you everything you need!