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Mathematical Induction problem

  1. Sep 29, 2005 #1
    I've been working on this problem for the past two and a half hours. The problem is: Prove by induction that n^3 + 2n is divisible by 3 for non-negative integer values of n.

    No matter how I try, I cannot manipulate the expression of the inductive step so that it is divisible by 3.

    I have done the base step and assumed that P(k): k^3 + 2k is divisible by 3. Now I need to show that this implies that P(k+1): (k+1)^3 + 2(k+1) is divisible by 3. I have tried manipulating P(k+1) in probably every possible way. The closest I have got to solving the problem, is getting part of the expression to be k^3 + 2k (which is divisible by 3 by assumption) and a constant 3.

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

    = k^3 + 2k^2 + 2k + 1 + 2k + 2

    = k^3 + 2k^2 + 4k + 3

    I also tried P(k+1) - P(k) and see if that would give an expression that is factorizable by 3. (Since the difference of two integers a and b is divisible by x, if and only if a and b are divisible by x.)

    P(k+1) - P(k)

    = [(k+1)^3 + 2(k+1)] - [k^3 + 2k]

    = [k^3 + 2k^2 + 2k + 1 + 2k + 2] - [k^3 + 2k]

    = 2k^2 + 2k + 3

    which does not lead to a desirable expression.

    I think I've exhausted all the options that I know of. Can anyone help me please? :cry:
  2. jcsd
  3. Sep 29, 2005 #2


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    Note that [itex](n+1)^3 + 2(n+1) = n^3 + 3 n^2 + 5 n + 3 = n^3 + 2 n + 3(n^2 + n + 1)[/itex]

    Therefore, if [itex]n^3 + 2n[/itex] is divisible by 3 then so is [itex](n+1)^3 + 2(n+1)[/itex]
  4. Sep 29, 2005 #3


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    P.S. This is not by induction but I noticed that

    [tex]n^3 + 2 n = n^3 + 3 n^2 + 2 n - 3n^2 = n(n+1)(n+2) - 3n^2[/tex]

    Since n is an integer then one of n, n+1 and n+2 must be divisible by 3 as is [itex]3n^2[/itex] so the original expression is divisible by 3 as well.
  5. Sep 29, 2005 #4
    Arrrrgh damn it, no wonder I couldn't get it right. I got the binomial expansion wrong! Thanks a lot, Tide, and that's a neat second proof. Can't believe I wasted a couple hours due to a stupid slip-up...
    Last edited: Sep 29, 2005
  6. Sep 29, 2005 #5
    Another one I absolutely could not do (and this time I'm sure it's not due to fumbling algebra) is this:

    Prove by induction that (1 + 1/4 + 1/9 + ... + 1/n^2) is less than (2 - 1/n) for all integers greater than 1.

    I get the base step, then make the assumption p(k) is true. Then I add 1/(k+1)^2 to both sides of the inequality.

    (1 + 1/4 + 1/9 + ... + 1/k^2 + 1/(k+1)^2) < 2 - 1/k + 1/(k+1)^2

    I guess the desired result is to prove that the RHS is less than 2 - 1/(k+1), but I cannot achieve that at all. This one seems definitely more difficult than the last one. Can anybody help me please?
  7. Sep 30, 2005 #6


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    Does this help?

    [tex]-\frac {1}{k} + \frac {1}{(k+1)^2} < -\frac {1}{k} + \frac {1}{k+1}[/tex]
  8. Sep 30, 2005 #7
    Thanks a lot, Tide. Actually I got something similar to that, and THEN gave up, not sure how to proceed. Induction sure is a great way to get stuck :(
  9. Sep 30, 2005 #8


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    Actually, induction is the easy part! Someone else did the grunt work of figuring out the appropriate inequality in the first place.
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