Question

Prove that \[n ^{3}-9n ^{2}+20n\] is divisible by 6, for all integers n≥1?

Answer 1

I used the wolfram|alpha website... but still a little confused

http://www.wolframalpha.com/input/?i=%28n%5E3-9n%5E2%2B20n%29%2F6

Answer 2

something about induction... and all that :)

Answer 3

p(1)=\[1^{3}-9(1)^{2}+20(1)=\] 12 and 12 is divisible by six...

Answer 4

Yes. Now suppose it's true for n:

Prove for n+ 1

Answer 5

ao therefore we can assume p(k) is true.

Answer 6

Assuming that is it true for n=k\[p(k)=k ^{3}-9k _{2}+20k\]?

Answer 7

The above was written incorrectly :)

\[p(k)=k ^{3}-9k ^{2}+20k\]

Answer 8

Now prove for n=k+1

Answer 9

If we let n = k+1,
(k+1)^3 -9(k+1)^2 +20(k+1)
=(k^3 +3k^2 +3k +1) -(9k^2+18k+9) +(20k+20)
=k^3 -6k^2+5k+12

Answer 10

\[p(k+1)=k+1^{3}-9(k+1) ^{2}+20(k+1)\]

Answer 11

oh, wait, someone already did it... thanks :)... oops

Answer 12

Now you must try to force the the divisibility

Answer 13

hmm im stuck as to how to prove that is divisible by 6
if we can show it is both divisible by 2 and 3, that would work

Answer 14

\[k^3-6k^2+5k+12 = (k^3-9k^2+20k)+3k^2-15k+12\]by inductive hypothesis, whats in parenthesis is divisible by 6. so we need to show that 3k^2-15k+12 is also divisible by 6. we obtain:\[3k^2-15k+12 = 3(k^2-5k+4) = 3(k-4)(k-1)\]

Answer 15

someone can check my quick math here

\[(k+1)^3 -9(k+1)^2 +20(k+1)=k^3-9k^2+20k+3(k-4)(k-1)\]

Answer 16

one of k-4 and k-1 will always be even, so you can pull a 2 from it, and coupled with the 3, will make 6.