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# Stuck on proof by induction question watch

1. Given that n is a positive integer, greater than 2, use the method of mathematical induction to prove that 2n>2n

I've shown that its true for the base step n=3.

Assume true for n=k
=>2k>2k
=>2k+1>2(k+1)

Not sure what to do next
2. (Original post by bobbricks)
Given that n is a positive integer, greater than 2, use the method of mathematical induction to prove that 2n>2n

I've shown that its true for the base step n=3.

Assume true for n=k
=>2k>2k
=>2k+1>2(k+1)

Not sure what to do next

I do not understand your structure

Assume true for n=k
=>2k>2k
=>2k+1>4k =2k+2k>2k + 6 > 2k +2 ....
3. (Original post by TeeEm)
I do not understand your structure

Assume true for n=k
=>2k>2k
=>2k+1>4k =2k+2k>2k + 6 > 2k +2 ....
Where did the 4k come from?
Assume true for n=k
=>2k>2k

Then for n=k+1
=>2k+1>2k+2
4. (Original post by bobbricks)
Where did the 4k come from?
multiplied both sides by 2
5. (Original post by bobbricks)
Then for n=k+1
=>2k+1>2k+2
This is what you are trying to prove. You cannot just state it. You need to use to prove that .

You have only assumed that it is true for n=k, not for n=k+1.
6. Assume true for n=k
=> 2k > 2k

Then for n=k+1
=> 2k+1 = 2*(2k) > 2*2k [from above]

=>
2k+1 > 2k + 2k > 2k + 2 [as k > 2]

And so on...

Key part is highlighted in bold

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