Induction (1 Viewer)

[phoenix159]

Prove by induction that n(n + 1)(n + 2) is divisible by 6

When n = 1, n(n + 1)(n + 2) = 6, which obviously works

Assume true for n = k, i.e. k(k + 1)(k + 2) = 6M

Consider n = k + 1

(k + 1)(k + 2)(k + 3) = 6M/k (k + 3) <--- how do i prove this is divisible by 6?

 

[Ikki]

Lol bro ur done, the six is factored out, I'm pretty sure u can say:
Since i factored a 6, it is divisible by 6

 

[phoenix159]

6M/k (k + 3) = 6 [M/k][k + 3]
But M/k might not be a whole number

 

[Ikki]

Hmm, I see your point. When all else fails, try a traditional expansion...

Step 3: Prove true for n=k+1

Assumption:


Proof:







Looks like you're right, it may not be divisible by 6 (maybe 3?)... Someone correct me? lol

 

[bottleofyarn]

Hmm, I see your point. When all else fails, try a traditional expansion...

Step 3: Prove true for n=k+1

Assumption:


Proof:







Looks like you're right, it may not be divisible by 6 (maybe 3?)... Someone correct me? lol

Expansion is the way to go. Usually with this question and its variants, there is a previous part where you explain why k(k+1) is even.

 

[RealiseNothing]

The LHS is just the sum of the first 'n' triangular numbers, so it is an integer. Thus the RHS is also an integer and hence is divisible by 6.

Who needs induction anyway?

 

[RealiseNothing]

Hmm, I see your point. When all else fails, try a traditional expansion...

Step 3: Prove true for n=k+1

Assumption:


Proof:







Looks like you're right, it may not be divisible by 6 (maybe 3?)... Someone correct me? lol

Now either k+1 or k+2 is even, so we can factor out a 2, and combined with the 3 we get 6. So a 6 can be factored out.

 

[Kurosaki]

6M/k (k + 3) = 6 [M/k][k + 3]
But M/k might not be a whole number

Um, by assumption

Ergo,
But, since k is an integer, (k+1)(k+2) is an integer, and hence so will the RHS?
Did I make an assumption too soon or something?
If I did, sorry: I suck.

 

[SpiralFlex]

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Edit: I realised I posted this yesterday in a thread. But using a combinatorical argument to prove that expression/6 was an integer.


Note: In conclusion if you are trying to prove divisibility of some consecutive numbers by some number A and cannot write it A*(integer). Try arguing even/divisible by 3 and let the appropriate term(s) = 2k or 3j

 

[RealiseNothing]

Um, by assumption

Ergo,
But, since k is an integer, (k+1)(k+2) is an integer, and hence so will the RHS?
Did I make an assumption too soon or something?
If I did, sorry: I suck.

Who says k divides M?

 

[SpiralFlex]

It does if the statement is in fact true. (which it is)

 

[RealiseNothing]

It does if the statement is in fact true. (which it is)

Cyclic argument.

 

[SpiralFlex]

^Yes

 

[Ikki]

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.
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.
.
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Edit: I realised I posted this yesterday in a thread. But using a combinatorical argument to prove that expression/6 was an integer.


Note: In conclusion if you are trying to prove divisibility of some consecutive numbers by some number A and cannot write it A*(integer). Try arguing even/divisible by 3 and let the appropriate term(s) = 2k or 3j

Can always rely on good ol' spiral. And sy LOL. U guys are crazy. lol
That's pretty cool...
I have to remember that you can do this, i've done a fair few inductions and this one's pretty special.

 

[Trebla]

Another approach:

(k + 1)(k + 2)(k + 3)
= k(k + 1)(k + 2) + 3(k + 1)(k + 2)
= 6M + 3(k + 1)(k + 2)

But recall that for integer k we have the arithmetic series
1 + 2 + 3 + .... + k + (k + 1) = (k + 1)(k + 2)/2
The LHS is obviously an integer so we can call it say N hence

(k + 1)(k + 2) = 2N

so the final expression becomes

= 6M + 6N
= 6(M + N)

Hence divisible by 6