Mathematical Induction Question (1 Viewer)

LostAuzzie

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Could someone please help me out with this question
Im stuck with how to write the answer out without making any assumptions no matter how obvious

Question:
Prove by mathematical Induction, n^3 > 2n^2 + n for all n>2

Your help would be much appreciated
 

rama_v

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LostAuzzie said:
Could someone please help me out with this question
Im stuck with how to write the answer out without making any assumptions no matter how obvious

Question:
Prove by mathematical Induction, n^3 > 2n^2 + n for all n>2

Your help would be much appreciated
I dunno if this is right but could you do this?
n=k+1
Need to prove (k+1)3 - 2(k+1)2 -(k+1) >0

LHS = (k+1)((k+1)2-2(k+1) -1)
= (k+1)((k2 + 2k +1) - 2k -2 -1)
= (k+1)((k2 -2))
but k>2 so (k+1)>0..and k2 -2 >0 when k>2 also (draw a parabola and this becomes clear)

.: LHS = (k+1)((k2 -2)) >0

therefore true for n=k+1
 

haboozin

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this is a classic question, they tend to not put inequalities in 3u for induction because there is only a few of them and they can almost be memorised.

i like the way rama v solved it, im gonna use this method next time.
i hate the way my teacher does it.
 

Raginsheep

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haboozin said:
this is a classic question, they tend to not put inequalities in 3u for induction because there is only a few of them and they can almost be memorised.

i like the way rama v solved it, im gonna use this method next time.
i hate the way my teacher does it.
Correct me if Im wrong but this is a math induction question and thus you need to make and use an assumption. Proving it the way above is, although mathematically correct, not maths induction.
 

rama_v

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Raginsheep said:
Correct me if Im wrong but this is a math induction question and thus you need to make and use an assumption. Proving it the way above is, although mathematically correct, not maths induction.
lol yeah, i just skipped to the n=k+1 step. Technically it shold start off with n=3, then assume true for n=k and then my step (prove true for n=k+1). Just add the first two steps and then its true by mathematical induction...
 

rama_v

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LostAuzzie said:
Could someone please help me out with this question
Im stuck with how to write the answer out without making any assumptions no matter how obvious

Question:
Prove by mathematical Induction, n^3 > 2n^2 + n for all n>2

Your help would be much appreciated
Hey mate u want to do nano technology too :D
I want do do that as well, at UTS, I'll probably be seeing you there next year
 

thunderdax

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rama_v said:
lol yeah, i just skipped to the n=k+1 step. Technically it shold start off with n=3, then assume true for n=k and then my step (prove true for n=k+1). Just add the first two steps and then its true by mathematical induction...
Er, no its not. To prove by mathematical induction you have to use the assumption.
 

mattchan

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The assume statement is:
k^3 - 2k^2 - k > 0


Expand (k+1)^3 - 2(k+1)^2 -(k+1) >0 to get:

= k^3 - 2k + k^2 - 2
= k^3 - 2k^2 - k + (3k^2 - k - 2) - Input the Assume Statement and Fix it up
= k^3 - 2k^2 - k + (k - 1)( 3k + 2)

Therefore for above line:
k^3 - 2k^2 - k > 0
(k - 1)( 3k + 2) > 0 for k > 1 and k <-2/3. Thus for k> 2, (k - 1)( 3k + 2) > 0

Therefore inequality is true for n = k + 1
 
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rama_v

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thunderdax said:
Er, no its not. To prove by mathematical induction you have to use the assumption.
oh right, yep, sorry i didnt fully understand ur first point. Anyway me mate chan solved it above :D
 

LostAuzzie

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rama_v said:
Hey mate u want to do nano technology too :D
I want do do that as well, at UTS, I'll probably be seeing you there next year
Most likely so, unless I have a sudden urge to travel for a year
An unlikely occurrence but a possibility nonetheless

Anyway thanks everyone for your help, much appreciated :uhhuh:
 

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