Question

My conjecture is n^3 - n is divisible by 6. I have done the basis step, and now have to do the inductive step, and i'm kinda stuck in that. Help??

Answer 1

so now
you want to prove that n(n+1)(n+2) is divisible by 6 , right ??

Answer 2

well did u write equation in k ?

Answer 3

yes, and yes :P

Answer 4

@hartnn n(n+1)(n-1) *

Answer 5

take 2 cases, n odd
n even

Answer 6

whats that shubham ?

Answer 7

this is where i'm at right now:

\[k^3 -6m(k+1)(k+2)\]

Answer 8

hmm...how ? :P

Answer 9

n^3 - n = n(n^2 -1) = n(n+1)(n-1)
how did you write n(n+1)(n+2) ?

Answer 10

i thought using induction, u assume the result for 'n'
and prove it for 'n+1'

Answer 11

@tanvidais13 can u show hw u gt to that?

Answer 12

because k was outside, and so then I replaced it with k^3 + 6m

Answer 13

the case when we have to show the result for 'n+1'
(n+1)^3-(n+1)= n(n+1)(n+2)

Answer 14

because k^3 - k = 6m, where m is any integer (showing that it is divisible by 6)

Answer 15

ideally, i would have liked to get 6m(something), and therefore it is divisble by 6. Right??

Answer 16

OH WAIT OMG I JUST GOT IT

Answer 17

got it ? good :P

Answer 18

yep infact use
k^3 = 6m+k

Answer 19

no.

Answer 20

then?

Answer 21

because i got the k(k+1)(k+2) thing, but how do I show that its divisble by 6??

Answer 22

i didn't know my presence have that effect!
ppl get answers on their own.....

Answer 23

lols @hartnn

Answer 24

ohh... take 2 cases
k even
k odd

Answer 25

oh but lulz.

Answer 26

but i can't do that. I have to 'induct' *nods head wisely*

Answer 27

when k= odd, k+1 is even
and either k or k+2 is divisible by 3.

Answer 28

you already used induction and now you want to prove k(k+1)(k+2) is divisible by 6, anyhow.....

Answer 29

^yup.

Answer 30

when k= even, k+1 is odd
and either k+1 or k+2 is divisible by 3.
hence in either case....

Answer 31

i just got something ,well we have

\[k(k+1)(k-1)=6m\]

we hav to prove

\[k(k+1)(k+2) \text{ is divisible by 6 }\]

then why dont u substitute

\[k(k+1)\]

from first equation?

Answer 32

isnt that direct method ?

Answer 33

wait i don't get it.

Answer 34

@hartnn what do you say?

Answer 35

but i've already substituted k^3 for 6m + k

Answer 36

@tanvidais13 mine is a fresh attempt by me dont confuse with ur working

Answer 37

oh ok.

Answer 38

did u understand it now @tanvidais13 ?

Answer 39

but that means dividing. and that's painful.

Answer 40

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

Since this is in the form of 3(some stuff), and 6 is a multiple of 3, the k(k+1)(k+2) is divisible by 6.

I don't know if this works, just thought I'd try.

Answer 41

then*

Answer 42

@blurbendy by your metjod i can make any expression divisible by 6 it seems :P

Answer 43

ok, i think i've got the answer. It's a mixture of mine and Arvind's method, but ok :P Anyhoo, thanks guyss :D

Answer 44

:) seems ?

Answer 45

and ok, stay on openstudy for a while, 'cuz there are going to be some more questions coming your way,

Answer 46

yw ... :) post them separately ..we will try :)

Answer 47

ok yay :D

Answer 48

TIME for me and pikachu to carry on with our adventures lols