Final answer:
The statement can be proven using mathematical induction. We prove the base case and then use the inductive hypothesis to prove the inductive step. By rearranging terms and factoring out a common factor, we show that the expression is divisible by 64. Therefore, the statement holds for every positive integer n.
Step-by-step explanation:
We will prove the statement using mathematical induction.
Base case: For n = 1, we have:
3^(2(1)+2) + 56(1) + 55 = 3^4 + 56 + 55 = 81 + 56 + 55 = 192.
Since 192 is divisible by 64, the statement holds for n = 1.
Inductive step: Assume the statement holds for some value of n: 3^(2n+2) + 56n + 55 is divisible by 64.
We need to prove the statement holds for n = k+1:
3^(2(k+1)+2) + 56(k+1) + 55 = 3^(2k+4) + 56k + 111.
Using the inductive hypothesis, we can write this expression as:
(3^2 * 3^(2k+2)) + 56k + 111.
By rearranging terms, we have:
(9 * 3^(2k+2)) + 56k + 111.
Factoring out 9, we get:
9 * (3^(2k+2) + (56k + 111)/9).
Since the second term is an integer (56k + 111) divided by 9, we can replace it with another integer without affecting the divisibility by 64.
Therefore, we have:
9 * (3^(2k+2) + m), where m is an integer.
Thus, the expression is divisible by 64.
Hence, by mathematical induction, we have proved that 64 divides 3^(2n+2) + 56n + 55 for every positive integer n.