Non-Standard Mathematical Induction Recurrences

When performing a proof by induction, we have typically applied a standard approach where:

  1. The base case was n=0n = 0 or n=1n = 1, and
  2. The inductive step was to prove that a result holds for n=k+1n = k+ 1, under the hypothesis that it holds for n=kn = k

However, there are many other formulations of problems where this approach would not work, but we can still use an inductive approach. For example, we may be asked to

Prove the n2+2nn^2 + 2n is divisible by 88 for any positive even integer nn

To achieve this, we can still use a base case of n=0n = 0, since 00 is indeed an even integer, but now our inductive step will be to show that the proposition holds for n=k+2n = k + 2, under our inductive hypothesis that it holds for n=kn = k. If we can prove this inductive step, then

  • If we can prove the base case, we know the proposition holds for n=0n = 0, then
  • By our inductive step, we know the proposition holds for n=0+2=2n = 0 + 2 = 2, then
  • By our inductive step, we know the proposition holds for n=2+2=4n = 2 + 2 = 4, then
  • By our inductive step, we know the proposition holds for n=4+2=6n = 4 + 2 = 6,
  • and so on

By doing this, we are proving the proposition for each even number one by one and so we prove the proposition for all even integers in this way.

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