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Mathematical induction problem.

  1. Jul 24, 2011 #1
    Hi. I am learning mathematical induction on my own and I came across this problem:

    1. The problem statement, all variables and given/known data
    Prove:
    1*4 + 4*7 + 7*10 + ... + (3n - 2)(3n + 1) = n(n + 1)²


    2. The attempt at a solution

    Quick test for n=1:

    (3 -2)(3 + 1) = 1(1 + 1)²
    4 = 4

    Alright, so I rewrite this with, on the left side, after the '...' having two members:

    1*4 + 4*7 + 7*10 + ... + (3n - 5)(3n - 2) + (3n - 2)(3n + 1) = n(n+1)²

    Assume n = k and it stands for k:

    1*4 + 4*7 + 7*10 + ... + (3k - 5)(3k - 2) + (3k - 2)(3k + 1) = k(k+1)²

    Lets prove that it stands for n = k+1:

    [1*4 + 4*7 + 7*10 + ... + (3k - 2)(3k + 1)] + (3k+1)(3k+4) = (k+1)(k+2)²

    Now replace everything in brackets with k(k+1)² from the step above:

    k(k+1)² + (3k+1)(3k+4) = (k+1)(k+2)²

    k³ + 2k² + k + 9k² + 12k + 3k + 4 = (k+1)(k² + 4k + 4)

    k³ + 11k² + 16k + 4 = k³ + 4k² + 4k + k² + 4k + 4 | -k³

    11k² + 16k + 4 = 5k² + 8k + 4

    6k² + 8k = 0


    Where's the mistake?
     
  2. jcsd
  3. Jul 24, 2011 #2

    SammyS

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    Does it work for n = 2? No!

    Added in Edit:

    Try: 1*4 + 2*7 + 3*10 + 4*13 + ... (n)(3n+1) = n(n+1)2 .
     
    Last edited: Jul 24, 2011
  4. Jul 24, 2011 #3
    It's not a true statement. That just goes to show you the proof by induction can be counted on :)
     
  5. Jul 24, 2011 #4
    Now it is true. Oh, and should I keep testing values like up to n=3 or something? Because everywhere I look they just say that you have to test for n=1.
    So is my book lying? Because it asks to "Prove that these equalities are right/true"
    Or is it the lack of my English skills and I didn't completely understand your post?
     
  6. Jul 24, 2011 #5
    By "not a true statement" I meant the left side of the equation did not equal the right side. You fixed it, now it does.

    You don't have to keep testing for higher values than 1, unless the equality is only true for something starting higher up, like, it's true for n>6 or something, then you test it for the first true value and then do k+1.
     
  7. Jul 24, 2011 #6
    I still am missing something here... How could it be not a true statement if they write in the book that those equalities are true? I solved a supposed to be right equality and it showed up that it wasn't right. The exercise should then ask to Prove whether equalities are right or not.
     
  8. Jul 24, 2011 #7
    I just meant 1*4 + 4*7 + 7*10 + ... + (3n - 2)(3n + 1) = n(n + 1)² wasn't true for all n.
    1*4 + 2*7 + 3*10 + 4*13 + ... (n)(3n+1) = n(n+1)^2 is true.
     
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