A community for students.
Here's the question you clicked on:
 0 viewing
plitter
 3 years ago
How do I prove that \(n^3n\) is always divisible with 6? I sort of see the solution since \(n(n1)(n+1)\) will always have a part that is a 2 and a part that is 3. But I can't formulate it mathematically.
plitter
 3 years ago
How do I prove that \(n^3n\) is always divisible with 6? I sort of see the solution since \(n(n1)(n+1)\) will always have a part that is a 2 and a part that is 3. But I can't formulate it mathematically.

This Question is Closed

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.13(2n) is always divisible by 6 ... not that it helps

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the proof might be induction tho

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, induction is the best way.

plitter
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm, give me a sec and I will try.

plitter
 3 years ago
Best ResponseYou've already chosen the best response.0First \(f(n) = \frac{n^3n}{6}\) \[f(1) = \frac{0}{6} = 0\] so valid, then \(f(k)\) and \(f(k+1)\) \[f(k+1) = \frac{(k+1)^3(k+1)}{6}\] which boils down to \[f(k+1) = \frac{k(k+1)(k+2)}{6}\] which basically is \[f(k+1) = \frac{l^3l}{6} \] where \(k=l+1\), is that good enough?

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.2I'll prove it for \(n\ge 0\). Let \(P(n)\) be the statement that \(n^3n\) is divisible by \(6\). \(P(0)\) is obviously true as you said. Now assume that \(P(k)\) is true, that is \(k^3k\) is divisible by \(6\) then for P(k+1): \[(k+1)^3(k+1)=(k+1)(k^2+2k)=k^3+3k^2+2k=k^3k+3k(k+1).\] By the induction hypothesis \(k^3k\) is divisible by \(6\) and it's obvious that \(3k(k+1)\) is also divisible by \(6\) since either \(k\) or \(k+1\) is divisible by \(2\). Therefore \(P(k)\) implies \(P(k+1)\) and thus \(P(n)\) is true \(\forall n\ge 0\).

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.2You can easily show that it's also true for \(n<0\) since [call \(f(n)=n^3n\) ] \(f(n)=(n)^3+n=(n^3n).\)

plitter
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not sure about the \((k+1)^3(k+1) = (k+1)(k^2+2k)\) of your equations, but the rest is right and with less work :) but I hope you approve of my method as well.

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.2\[(k+1)^3(k+1)=(k+1)((k+1)^21)=(k+1)(k^2+2k).\] As for your method, you didn't state clearly what is the induction hypothesis and you didn't show that f(k) implies f(k+1). That's at least what I think.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.