1.1. Let P(n) be the claim that 3 | n(n + 1)(n + 2) for
all integers n1.
P(1) is the statement: 3 | 1.
P(k) is the statement: 3 | k(k + 1)(k + 2), or equivalently, k(k + 1)(k + 2) = 3a for some integer a.
P(k+1) is the statement: 3 | (k + 1)(k + 2) (k + 3), or equivalently, (k + 1)(k + 2)(k + 3) = 3b for some integer b.
For a proof by induction, you first need to check that the statement P(1) is true. Then assume that P(k) is true and use this to show that P(k + 1) is true.
The statement P(1) is true since 1 = 6 and 3 | 6.
Now assume that the statement P(k) is true. We now need to show that the left-hand side of P(k+1) is equal to the right-hand side of P(k+1). We shall use the second version of each of P(k) and P(k+1) since equations are easier to work with than statements involving the divides symbol.
| L.H.S. of P(k+1) | = | (k + 1)(k + 2)(k + 3) |
| = | (k + 3)(k + 1)(k + 2) | |
| = | k(k + 1)(k + 2) + 3(k + 1)(k + 2) | |
| = | 3a + 3(k + 1)(k + 2) (since we are assuming that P(k) is true) | |
| = | 3 [a + (k + 1)(k + 2)] (now a + (k + 1)(k + 2) is an integer since a and k are both integers) | |
| = | 3b (for some integer b) | |
| = | R.H.S. of P(k+1) |
Therefore, for all integers n1,
3 | n(n + 1)(n + 2).