Mathematical inductiuon help please (1 Viewer)

[mstellatos]

Could some one help me with this mathematical induction question

2n^3-2n is divisible by 12 for integers n> and equal to 1

 

[ellen.louise]

Could some one help me with this mathematical induction question

2n^3-2n is divisible by 12 for integers n> and equal to 1

for n=1
LHS=2(1)^3 - 2(1)
LHS=2 - 2 = 0???

This doesn't work: is there meant to be brackets around your '3 - 2n'???
make sure there's always brackets when there needs to be. Confirm and ill give it a shot!

 

[vulgarfraction]

And why wouldn't 0 be divisible by 12?

Umm... you can do it without induction.

But if you have to do it with induction:

let f(x)=2n^3-2n
say 12|f(k). then show 12|{f(k+1)-f(k)}

f(k+1)-f(k)= [you can expand this yourself] = 6k(k+1)

rtp 2|k(k+1) which is true because k and k+1 are opposite parity.

And you have the base case is true.

EDIT: And it can't possibly be true for 2n^(3-2n) because then that implies 6|n^(3-2n). If n is prime then you're immediately screwed.

 

[ianc]

okay these divisibility inductions are always a little dicey, but here goes:
(btw i assumed it to be 2n³ - 2n)


Show true for n=1:
2(1) - 2(1)
= 0
= 12(A) (where A = 0)
Therefore true for n=1

Assume true for n=k:
2k³ - 2k = 12B (where B is any integer)

Show true for n=k+1:
2(k+1)³ - 2(k+1)
= 2k³ + 6k² + 4k
= (2k³ - 2k) + 6k² + 6k
= 12B + 6k(k+1) (from assumption, where B is any integer)
With 6k(k+1), either k or k+1 is even, so k(k+1) will be divisible by 2
So (2k³ - 2k) + 6k(k+1)
= 12B + 6(2)C (where C is any integer)
= 12(B+C)
Therefore divisible by 12

Therefore true for n=k+1, so by induction true for all n>=1.



Hope this helps!

 

[ellen.louise]

And why wouldn't 0 be divisible by 12?

fuck, you have a point there!

 

[mstellatos]

Thanks soo much ..it was the 6k(k+1) that I was stuck on

 

[ianc]

yeah, i had to think a little about that step......often for division or inequality inductions you have to think about the nature of the numbers you are dealing with. i find playing with numbers on my calculator can help me to see patterns.

but remember with induction questions, they are usually worth about 4 marks, and if you had just written it all out with the correct proof for n=1 and assumption for n=k, then you are pretty much guaranteed at least 2 marks.

 

[milton]

2n^3 - 2n = 2(n-1)n(n+1)
obviously, one of (n-1), n or (n+1) must be a multiple of 3, so the expression is divisble by 6

 

[SeDaTeD]

milton: You can go one better and also say at least one of (n-1), n or (n+1) is divisible by 2, so the entire expression is divisible by 12.