(i) If

(ii) For all positive integers

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$n\ge 5$ ,

$2}^{k}>{k}^{2$

(i) If

(ii) For all positive integers

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Sally Cresswell

Answered 2020-12-28
Author has **91** answers

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

${2}^{n+1}{3}^{2(n+1)}-1={2}^{n}\cdot 2\cdot {3}^{2n}-1$

$=18\cdot {2}^{n}{3}^{2n}-1$

$=17\cdot {2}^{n}{3}^{2n}+{2}^{n}{3}^{2n}-1$

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,${2}^{n+1}{3}^{2(n+1)}-1$ is diviseble by 17

Therefore, whenever n = k is true n = k+1 is true.

Therefore, it is true for all integers.

If

that is:

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:

Now, prove that the statement is true for n = k+1:

So, consider

Now, since

hence,

Therefore, whenever n = k is true n = k+1 is true.

Therefore, it is true for all integers.

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