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Homework Help: Mathematical Induction

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Use mathematical induction to prove the formula for every positive integer n.

    [tex]\sum_{i=1}^{5}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    I know this will be true because the RHS is just your standard sums of powers of integers to the fifth formula, but nevertheless I have to prove it...

    I know n=1 will work so I set n=k

    [tex]\sum_{k=1}^{5}k^5=\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]

    I take the RHS and set k=(k+1)


    [tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2(k+1)-1)}{12}[/tex]
    [tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]

    If the test works for all the terms then it must be correct for the first 5 terms so I can eliminate the summation and just have the [tex]k^5[/tex]

    so then I'd have

    [tex](k+1)^5+\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]
    [tex]=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]

    From here I'm not sure how to get that to equal this:

    [tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]
     
  2. jcsd
  3. Jul 10, 2010 #2
    Your first formula as written is incorrect. The sum from 1 to 5 is not different for every positive integer, n. Your upper index should be n and not 5.

    To use induction you must:
    1. Verify the base case.
    2. Assume that your case holds for some number, k.
    3. Show that it holds for k+1. (IF it is true for k, THEN it is true for k+1)

    Try adding [tex](k+1)^5[/tex] to both sides of the equation and manipulate the right side until you get the proper form.
     
  4. Jul 10, 2010 #3
    My first formula is written exactly as printed from my text book so I know it is correct. I know how to solve with induction my question is only one of algebra.

    I need to get

    [tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]

    The one on the left is where I subbed in k=(k+1) the one on the right is the equation I got from adding (k+1)5 to. I know they are equal because you can plug in a value and get the same number on both sides. How can you get these equations into the same form.
     
  5. Jul 10, 2010 #4

    LCKurtz

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    Failing anything clever, you could always expand and collect both numerators to see they are both:

    2k6+18k5+65k4+120k3+119k2+60k+12
     
  6. Jul 10, 2010 #5

    ehild

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    Homework Helper

    That is correct. Note that both sides have the common factor (k+1)^2, so you have to prove the simpler equality

    [tex]\frac{(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2 (2k^2+2k-1)+12(k+1)^3}{12}[/tex].

    After that you can follow LCKurtz suggestion to expand both sides and see if they are identical.

    ehild
     
  7. Jul 11, 2010 #6
    Alright, that works.

    Thanks.
     
  8. Jul 11, 2010 #7
    At the risk of sounding like a stickler/jerk, the formula is NOT correct, even if it is what is listed in the book. Typically, a formula in n will be given when summing from 1 to n, but that is summing from 1 to 5.
     
  9. Jul 11, 2010 #8

    ehild

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    You are right, it must be a typo. So the correct form is

    [tex]
    \sum_{i=1}^{n}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}
    [/tex]

    ehild
     
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