Question

Prove using mathematical induction that 3 divides n(n+1)(n+2). I have the first step for n=1, and then assume 3 divides k(k+1)(k+2), k being an integer. I'm now at the next step of trying to prove 3 divides (k+1)(k+2)(k+3) and am stuck.

Answer 1

how did you prove the first part :)

Answer 2

or are we just to assume the k(k+1)(k+2) is divisible by 3

Answer 3

I have 1(1+1)(1+2)=6 which divides into 3. And then k=n, for k(k+1)(k+2). The examples from my prof go from showing n=1 to assuming the statement for k, then plugging in k+1 for k and solving from there...

Answer 4

Assume n=k is valid and sub into n=k+1

Answer 5

if (k+1)(k+2)(k+3) is a scalar of k(k+1)(k+2); then id say it goes

Answer 6

how do I prove that it is a scalar?

Answer 7

dunno, havent done any induced proofs

k+1
k+2
----
k(k^2 +3k +2)
(k+1)(k^2 +3k +2) perhaps?

Answer 8

err, (k+3)(k^2 +3k +2)

Answer 9

I have that much so far. Basically I am trying to get that to reduce to k(k+1)(k+2) plus whatever is left which 3 should divide into. k(k+1)(k+2) is already assumed divide by 3 that

Answer 10

(k+1)(k+2)(k+3)=1/3 (3k+1)(k+2) (k+1)
proved

Answer 11

I'm not really following that.... How did you get that?

Answer 12

you want to show in third step that it is divisible by 3 that I have shown you

Answer 13

if you now now multiply (3k+1)(k+2)(k+1), whatever you will get it must be divisible by 3

Answer 14

because 1/3 is now thier factor

Answer 15

Ok, I see. I wasn't sure where the 1/3 came from. Thanks