Question

prove this statement through mathematical induction: (3^n) - 1 is divisible by 2.

Answer 1

There is at least one \(n\) such that\[\frac{3^n-1}{2}\in\mathbb Z\]We can test this by picking any positigve integer. Does \(\frac{3^{n+1}-1}{2}\in\mathbb Z\) also?\[\frac{3^{n+1}-1}{2}=\frac{3^n3-1}{2}=\frac{3}{2}3^n-\frac{1}{2},\left\{\frac{3^n}{2}=a+\frac{1}{2}\mid\forall n\because3^n\in\mathbb Z,\frac{3^n}{2}\notin\mathbb Z\right\}\]I think you can do it from here?

Answer 2

\(3^n-1\) is even. Why bother using induction?

Answer 3

Because sometimes when you do the same thing multiple times you get a good grade.

Answer 4

I see what you're doing, but it needs to be done using induction :/

Answer 5

This is the only type of question I love :P
For n=1
(3^n) - 1 = (3^1) -1 =2, which is divisible by 2
So, the proposition is true for n=1
Assume (3^k) - 1 is divisible by 2 for some positive integers k, that is (3^k) - 1 =2m, where m is a positive integer.

Then n=k+1
(3^(k+1)) -1
= 3 x (3^k) -1
= 3( 2m +1) -1
= 6m +3 -1
= 6m +2
= 2(3m+1), where 3m+1 is a positive integer
therefore (3^(k+1)) -1 is divisible by 9
therefore the proposition is also true for n=k+1
By principle of MI, the proposition is true for all positive integers n

Answer 6

If the question asks you to do so, it's difficult to ask why :S

Answer 7

thanks!
but I don't get this line: = 3 x (3^k) -1

Answer 8

That line may have looked like this:\[3^1*3^k-1=3*3^k-1\]

Answer 9

Thanks for the explanation @rchstudy I didn't realise that there was a reply. Sorry!!

Answer 10

no prob @Callisto