1. Since the equation we are going to use to prove this statement, that is, the equation br = 5br-1 - 6br-2, relates br to two previous terms in the sequence, we shall need two statements in our basis step.
2. Let P(n) be the claim that bn = 2n
for all integers n1.
Since the equation we are going to use to prove this statement, that is, the equation br
= 5br-1 - 6br-2, relates br to two previous terms in the
sequence, we shall need two statements in our basis step.
P(1) is the statement: b1 = 21 and P(2) is the statement:b2 = 22 .
P(i) is the statement: bi = 2i .
P(k) is the statement: bk = 2k .
For a proof by strong mathematical induction, you first need to check that all the statements in your basis step (P(1) and P(2)) are true. Then assume that P(i) is true for all values of i where 1 < i < k and use this to show that P(k) is true.
The statement P(1) is true since we are told in the problem that b1 = 2 = 21. The statement P(2) is true since we are told in the problem that b2 = 4 = 22.
Now assume that the statement P(i) is true, for all values of i between 1 and k, that
is, for 1 < i < k. We now need to show that the left-hand side of P(k) is
equal to the right-hand side of P(k). We are told in the problem that br = 5br-1
- 6br-2 for all integers r3, so we can use that in the rest of our proof.
L.H.S. of P(k) |
= | bk |
| = | 5bk-1 - 6bk-2 | |
| = | 5 · 2k-1 - 6 · 2k-2 (since we are assuming that P(i) is true for 1 < i < k) | |
| = | 2k-2 (5 · 2 - 6) | |
| = | 2k-2 · 4 | |
| = | 2k | |
| = | R.H.S. of P(k) |
Therefore, the sequence b1, b2, b3, ... is
a sequence defined in this problem can be described using the general formula bn
= 2n for all integers n1.