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Stuck on proof by induction question

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 :confused:
Reply 1
Avatar for TeeEm
TeeEm
19
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 :confused:



I do not understand your structure


Assume true for n=k
=>2k>2k
=>2k+1>4k =2k+2k>2k + 6 > 2k +2 ....
Reply 2
Avatar for bobbricks
bobbricks
OP
19
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
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by bobbricks
Where did the 4k come from?


multiplied both sides by 2
Reply 4
Avatar for james22
james22
16
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 2k?2k2^k?2k to prove that 2k+1>2(k+1)2^{k+1}>2(k+1).

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