For any integer n ≥ 1, let Pn be the statement that 3⋅4n+51 is divisible by 9.
Base case. The statement P1 says that 3⋅41+51=63 is divisible by 9, which is true.
Inductive Step. Fix k ≥ 1, and suppose that Pk holds, that is, 3⋅4k+51 is divisible by 9,
3⋅4k+51=9m,m∈Z,m≥7 It remains to show that Pk+1 holds, that is, 3⋅4k+1+51 is divisible by 9,
3⋅4k+1+51=4⋅(3⋅4k)+51==4⋅(3⋅4k+51)−4⋅51+51=
=4⋅9m−3⋅51=9(4m−17),m∈Z,m≥7Therefore Pk+1 holds.
Thus, by the principle of mathematical induction, for all n ≥ 1, Pn holds.
Therefore 3⋅4n+51 is divisible by 9 for all positive integers n.
Therefore 3⋅4n+51 is divisible by 3 for all positive integers n.
Comments
Leave a comment