Question

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

Abstract algebra
ANSWERED
asked 2020-12-27
Use Principle of MI to verify
(i) If \(\displaystyle{n}\in\mathbb{Z}\) is a positive ineger then \(\displaystyle{2}^{{n}}{3}^{{{x}{n}}}-{1}\) divisible by 17.
(ii) For all positive integers \(\displaystyle{n}\ge{5}\),
\(\displaystyle{2}^{{k}}{>}{k}^{{2}}\)

Answers (1)

2020-12-28
According to the given information it is required to prove the following using principle of mathematical induction.
If \(\displaystyle{n}\in\mathbb{Z}\) is a positive ineger then \(\displaystyle{2}^{{n}}{3}^{{{2}{n}}}-{1}\) divisible by 17.
that is: \(\displaystyle{2}^{{n}}{3}^{{{2}{n}}}-{1}={x}^{{n}}{9}^{{n}}-{1}={18}^{{n}}-{1}\) 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:
\(\displaystyle{k}\in{Z}\) is posisitve integer then \(\displaystyle{2}^{{k}}{3}^{{{2}{k}}}-{1}={18}^{{k}}-{1}\) is divisible by 17
Now, prove that the statement is true for n = k+1:
So, consider
\(\displaystyle{2}^{{{n}+{1}}}{3}^{{{2}{\left({n}+{1}\right)}}}-{1}={2}^{{n}}\cdot{2}\cdot{3}^{{{2}{n}}}-{1}\)
\(\displaystyle={18}\cdot{2}^{{n}}{3}^{{{2}{n}}}-{1}\)
\(\displaystyle={17}\cdot{2}^{{n}}{3}^{{{2}{n}}}+{2}^{{n}}{3}^{{{2}{n}}}-{1}\)
Now, since
\(\displaystyle{17}\cdot{2}^{{n}}{3}^{{{2}{n}}}\) is divisible by 17 and from assumption \(\displaystyle{2}^{{n}}{3}^{{{2}{n}}}-{1}\) is divisible by 17
hence, \(\displaystyle{2}^{{{n}+{1}}}{3}^{{{2}{\left({n}+{1}\right)}}}-{1}\) is diviseble by 17
Therefore, whenever n = k is true n = k+1 is true.
Therefore, it is true for all integers.
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