HSC Stuff > HSC Mathematics Extension 1
Induction Inequality
kemi:
Greetings! Just registered, as I was intrigued by the friendly vibes coming from these forums :)
Rather simple question:
Prove the cube of the sum of consecutive integers is divisible by 3, where n ∈ N
At the 'assumption' step the proof has already become blatant - I am able to factor out 3. For some reason I feel this is wrong, since the point of the assumption is to use it for the proof. Even upon reaching the 'proof' i.e. prove true for n = k+1, it seems the assumption is unnecessary? Maybe I have made an error in setting up the q. Help would be much appreciated :D
Thank you!
jamonwindeyer:
--- Quote from: kemi on January 13, 2017, 01:32:53 pm ---Greetings! Just registered, as I was intrigued by the friendly vibes coming from these forums :)
Rather simple question:
Prove the cube of the sum of consecutive integers is divisible by 3, where n ∈ N
At the 'assumption' step the proof has already become blatant - I am able to factor out 3. For some reason I feel this is wrong, since the point of the assumption is to use it for the proof. Even upon reaching the 'proof' i.e. prove true for n = k+1, it seems the assumption is unnecessary? Maybe I have made an error in setting up the q. Help would be much appreciated :D
Thank you!
--- End quote ---
Hey kemi! Welcome to the forums! Sending some good vibes your way :)
What you've found is correct - This is an example of a mathematical fact that is just self apparently true. This is reflected in the fact that your assumption is true. So although you can use induction to do it, it's blatantly unnecessary (as you've seen). You should still be able to do the steps, even though we know you don't really need to.
Assume \(k+k+1+k+2=3M\implies3k+3=3M\)
So \(k+1+k+2+k+3=3M+3=3(M+1)\)
That's the rough working. You can still do it - Even though you don't need to :)
RuiAce:
--- Quote from: jamonwindeyer on January 13, 2017, 01:41:48 pm ---Hey kemi! Welcome to the forums! Sending some good vibes your way :)
What you've found is correct - This is an example of a mathematical fact that is just self apparently true. This is reflected in the fact that your assumption is true. So although you can use induction to do it, it's blatantly unnecessary (as you've seen). You should still be able to do the steps, even though we know you don't really need to.
Assume \(k+k+1+k+2=3M\implies3k+3=3M\)
So \(k+1+k+2+k+3=3M+3=3(M+1)\)
That's the rough working. You can still do it - Even though you don't need to :)
--- End quote ---
CUBES, Jamon!!!
RuiAce:
(0 is the smallest natural number.)
________________
Also, this wasn't really an inequality induction
de:
I realise that this is supposed to be done with induction. However, I'll just mention a smoother method. You can just consider the equation modulo 3. Then you can sub 0,1,2 into the equation and since and and we are done.
*modulo means considering all numbers as their remainder when divided by 3.
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