 # Use Principle of MI to verify (i) If n in ZZ is a positive ineger then 2^n 3^(xn)-1 divisible by 17. (ii) For all positive integers n>=5, 2^k>k^2 hexacordoK 2020-12-27 Answered
Use Principle of MI to verify
(i) If $n\in \mathbb{Z}$ is a positive ineger then ${2}^{n}{3}^{xn}-1$ divisible by 17.
(ii) For all positive integers $n\ge 5$,
${2}^{k}>{k}^{2}$
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According to the given information it is required to prove the following using principle of mathematical induction.
If $n\in \mathbb{Z}$ is a positive ineger then ${2}^{n}{3}^{2n}-1$ divisible by 17.
that is: ${2}^{n}{3}^{2n}-1={x}^{n}{9}^{n}-1={18}^{n}-1$ is divisible by 17
For n = 1
18-1=17
17 is diviseble by 17
so, the statement is true for n=1
Let us assume that the given statement is true for n = k that is:
$k\in Z$ is posisitve integer then ${2}^{k}{3}^{2k}-1={18}^{k}-1$ is divisible by 17
Now, prove that the statement is true for n = k+1:
So, consider
${2}^{n+1}{3}^{2\left(n+1\right)}-1={2}^{n}\cdot 2\cdot {3}^{2n}-1$
$=18\cdot {2}^{n}{3}^{2n}-1$
$=17\cdot {2}^{n}{3}^{2n}+{2}^{n}{3}^{2n}-1$
Now, since
$17\cdot {2}^{n}{3}^{2n}$ is divisible by 17 and from assumption ${2}^{n}{3}^{2n}-1$ is divisible by 17
hence, ${2}^{n+1}{3}^{2\left(n+1\right)}-1$ is diviseble by 17
Therefore, whenever n = k is true n = k+1 is true.
Therefore, it is true for all integers.