Makayla R.

# Prove that for any integer n, with n greater than or equal to 1...

Using mathematical induction, prove that for any integer n, with n≥1, n(n2-1)(n+2) is divisible by 4.

## 2 Answers By Expert Tutors

By: Private Tutor - English, Mathematics, and Study Skills

Makayla R.

Thanks! One question, though-- I thought that for induction, we would assume that the inductive hypothesis is true (the statement "k(k2-1)(k+2) is divisible by 4" is true), and then show that if P(k) is true that P(k+1) is true. Is k+1 the same as using k-1?
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05/18/15 Stephanie M.

tutor
Hi Makayla,

That's a great question. The short answer is, yes, the following two methods are equivalent:

(a) Assume it's true for n, and prove it's true for n+1
(b) Assume it's true for n-1, and prove it's true for n

That's because both serve the same function. We just want to prove that, given that one case is true, we could prove the next case is true. I used (b), but you're absolutely free to use (a). It will likely look pretty much the same.

Sometimes, it's easier to use one method or the other. Pick the one you like best and use that most of the time, but remember that you can always try the other method if the expression gets too hard to work with!

Does that make sense?

Stephanie

P.S.

Here's method (a), briefly:

Assume it's true for n: n4 + 2n3 - n2 - 2n is divisible by 4

Prove it's true for n+1, which simplifies to: n4 +6n3 + 11n2 + 6n
That's equivalent to n4 + 2n3 - 2n3 + 6n3 - n2 + n2 + 11n2 - 2n + 2n + 6n
That's equivalent to (n4 + 2n3 - n2 - 2n) - 2n3 + 6n3 + n2 + 11n2 + 2n + 6n
That's equivalent to (n4 + 2n3 - n2 - 2n) + 4n3 + 12n2 + 8n
All of those expressions are divisible by 4, so the statement is true for the n+1 case.
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05/18/15

Makayla R.

That does make sense! Thank you so much!
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05/19/15 