Answer to Question #291334 in Real Analysis for Nikhil

Question

Using the principle of induction, prove that 64 is a factor of 3^(2n+2)- 8n-9 ∀ n∈N

Answer 1

Let $P(n)$ be the proposition that $64$ is a factor of $3^{2n+2}-8n-9, \forall n\in \N.$

Basis Step

$P(1)$ is true because $64$ is a factor of $3^{2(1)+2}-8(1)-9=64.$

Inductive Step

We assume that $64$ is a factor of $3^{2k+2}-8k-9, k\in \N.$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$64$ is a factor of $3^{2(k+1)+2}-8(k+1)-9, k\in \N.$

$$3^{2(k+1)+2}-8(k+1)-9=9\cdot3^{2k+2}-8k-17$$

$$=9\cdot3^{2k+2}-72k+64k-81+64$$

$$=9(3^{2k+2}-8k-9)+64(k+1)$$

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for every natural $n$. That is, we have proved that $64$ is a factor of $3^{2n+2}-8n-9, \forall n\in \N.$