Using PMI, prove that \[{{3}^{2n+2}}-8n-9\] is divisible by 64.

Hint: Put n = 1 and prove that the expression is divisible by 64. Then put n = k, where it is divisible by 64. Then prove that n = k + 1 is also true i.e. prove that P (k + 1) is true whenever P (k) is true.

Complete step-by-step answer:
PMI is the principle of Mathematical induction. This method is used for proving statements, a theorem or a formula that is asserted about a natural number.
Let us consider, \[P\left( n \right)={{3}^{2n+2}}-8n-9-(1)\].
We need to prove that P (n) is divisible by 64.
Let us first put n = 1 in equation (1).
\[\begin{align}
  & \therefore P\left( 1 \right)={{3}^{2+2}}-8-9={{3}^{4}}-8-9 \\
 & P\left( 1 \right)=81-8-9=64 \\
\end{align}\]
Therefore, P (n) is divisible by 64 when, n = 1.
Let us assume that n = k and we get,
\[P\left( k \right)={{3}^{2k+2}}-8k-9\], which is divisible by 64.
Let us consider that, \[{{3}^{2k+2}}-8k-9=64m\] where, \[m\in N\].
\[\because \] N = natural number.
\[{{3}^{2k+2}}-8k-9=64m-(2)\]
Now we need to prove that P (k + 1) is also true.
Put, n = k + 1.
\[P\left( k+1 \right)={{3}^{2\left( k+1 \right)+2}}-8\left( k+1 \right)-9\]
Thus simplify the above expression.
\[P\left( k+1 \right)={{3}^{2k+2+2}}-8k-8-9\]
Using product rule of exponent, split \[{{3}^{2k+2+2}}\] into \[{{3}^{2}}{{.3}^{2k+2}}\].
\[P\left( k+1 \right)={{3}^{2}}{{.3}^{2k+2}}-8k-17-(3)\]
From equation (2) find the value of \[{{3}^{2k+2}}\].
\[{{3}^{2k+2}}=64m+8k+9\]
Substitute this value in equation (3).
\[\begin{align}
  & P\left( k+1 \right)=9\left( 64m+8k+9 \right)-8k-17 \\
 & P\left( k+1 \right)=9\times 64m+72k+81-8k-17 \\
 & P\left( k+1 \right)=9\times 64m+64k+64 \\
\end{align}\]
\[P\left( k+1 \right)=64\left( 9m+k+1 \right)\], which is divisible by 64.
Thus we proved that P (k + 1) is true whenever P (k) is true. Hence by principal mathematical Induction; P (x) is true for all natural number, \[P\left( x \right)={{3}^{2n+2}}-8n-9\] is divisible by \[64 n\in N\].

Note: We said that PMI is a method for proving statements of the form P (n). It works because of how the natural numbers are constructed from set theory. PMI is built into N (natural number). PMI is a kind of deductive argument, a logically vigorous method of proof.