Question

can sum1 help me w/ pre calc?

Answer 1

Use mathematical induction to prove that 2 is a factor of (n+1)(n+2) for all positive integers "n"

Answer 2

Prove it for the base case n=1, then prove the induction step, assume true for n, prove for n+1.

Answer 3

then give an alternate proof of the assertion in the previous problem based on the fact that (n+1)(n+2) is a product of two consecutive integers

Answer 4

agh can you show me @nbouscal ?
i havent done this since last semester :/

Answer 5

The alternate proof will rely on the fact that one of two consecutive integers will always be even, I believe.

Answer 6

It's simple. First, show for n=1. So, (1+1)(1+2)=(2)(3)=6, 2 is a factor of 6. Next, induction step. We assume that (n+1)(n+2) is even (another way of saying 2 is a factor if it). So we show that it's true for n+1, namely, (n+2)(n+3) is even. Let's multiply those both out. We have for the first one, n^2+3n+2, and for the second, n^2+5n+6. We know that n^2+3n+2 is even from the induction assumption, and we have the n+1 case is just that plus an extra 2n+4. 2n+4=2(n+2), so it's even, and an even plus an even is even.

I didn't write that very formally but that's the idea.

Answer 7

w8 why do we assume it is even?

Answer 8

waittt
oh nvm i get it

Answer 9

thx :D

Answer 10

best way:Between two consecutive numbers is a one even.