1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial theorem-related proof

  1. Sep 13, 2014 #1
    I've been thinking about this for some time. Now I'm coming on here in the hope of getting some help;

    Prove that an+3 + (a + 1)2n+3 is divisible by a2 + a + 1.

    I can't quite remember the restrictions on n, though I'd imagine it'd be "for all real n ≥ -1" or something similar.

    Thanks in advance! :)

    EDIT: The way to do this would probably be through the method of mathematical induction, since that was the subject of class the day this problem was given.
     
    Last edited: Sep 13, 2014
  2. jcsd
  3. Sep 13, 2014 #2

    Char. Limit

    User Avatar
    Gold Member

    Mathematical induction is definitely the way to go, yeah. Proving that it's true for n=-1 is rather simple, as I'm sure you've already seen. Then you just need to take the expression:

    [tex]a^{(n+1)+3} + (a + 1)^{2(n+1)+3}[/tex]

    And rewrite it into an expression of the form:

    [tex]f(a) \left(a^{n+3} + (a+1)^{2n+3}\right)[/tex]

    Where f(a) can be any function. Since you know the latter is divisible by [itex]a^2+a+1[/itex] by assumption, then the full expression will be divisible by [itex]a^2+a+1[/itex] as well, and that will conclude the proof!
     
  4. Sep 13, 2014 #3
    Can I really assume that an+3 + (a + 1)2n+3 is divisible by a2 + a + 1, when that is what I am aiming to prove?

    Sorry, just trying to get into the whole induction thinking...

    Thanks for your time, btw :) really appreciate it!
     
  5. Sep 13, 2014 #4

    Char. Limit

    User Avatar
    Gold Member

    Yes, you can. In fact, that's the whole point of induction! You prove a certain base case (here, n=-1), then you make the assumption that it's true for an unspecified number n, and using that assumption, prove that it's also true for n+1. Both of those facts together show that it's true for all n≥-1, as long as n's an integer!
     
  6. Sep 13, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, that's NOT what you are trying to prove. What you are trying to prove is that "an+3 + (a + 1)2n+3 is divisible by a2 + a + 1 for all n". What you are assuming with induction is "an+3 + (a + 1)2n+3 is divisible by a2 + a + 1 for a single value of n.
    (I prefer to use a different letter, say "k", rather than "n" when stating induction hypothesis to avoid that mistake: "if a statement is true for n= 1 (or 0 or -1) and whenever the statement is true for a number, k, it I also true for k+1 then the statement is true for all n.")

    That is, once you have proved it true for -1, you then show it is true for =1+1= 0. Now that you know it is true for 0, you show it is true for 0+1= 1. Now that you know it is true for 1, you show it is true for 1+ 1= 2.

    Showing that "if it is true for k then it is true for k+1" collapses all of those into one proof.
     
  7. Sep 13, 2014 #6
    Ah, yes, I shall think about that in the future.

    As for this problem, though. I still can not solve it :(
     
  8. Sep 13, 2014 #7
    I'm going to need some hint, I've tried some different ways; Writing (a + 1)^(2n+3) as a sum, futile long division.
    Thanks in advance.
     
  9. Sep 24, 2014 #8
    In = an+3 + (a+1)2n+3 , when n=-1 and n=0 it can be shown that it is divisible by a2+a+1
    In+1 = an+4 + (a+1)2n+5
    In+1 - In = (an+4 - an+3) + [(a+1)2n+5 - (a+1)2n+3]
    which can be ultimately reduce to In+1 = a.In + (a2+a+1).(a+1)2n+3
    So, by induction method u can prove. Not true for n less than -1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Binomial theorem-related proof
  1. Binomial Theorem (Replies: 1)

  2. The Binomial Theorem (Replies: 2)

  3. Binomial Theorem (Replies: 6)

  4. Binomial Theorem (Replies: 3)

Loading...