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Homework Help: Proof by induction

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove by induction, that for all integer n where n>= 1

    [tex] \sum_{i=1}^{n} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4} [/tex]



    2. Relevant equations



    3. The attempt at a solution

    First question is do I start at i=0 or i=1? It says >=, so not sure.

    Ok then I added (n+1)(n+2)(n+3) to the right to balance out the n+1 on the left. But what do I do after that. Just simplify? Or am I missing something

    [tex] \sum_{i=1}^{n+1} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3) [/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 28, 2010 #2
    Just add (n+1)(n+2)(n+3)=x to the fraction by multiplying it by 4/4, and factor x out of the resulting terms.
     
  4. Feb 28, 2010 #3
    Ok, so basically just plugin n+1 for i on the left, and simplify the right. Set them equal, and done?
     
  5. Feb 28, 2010 #4
    You went to n+1 in your sum, took out the n+1 case and applied the induction hypothesis to the sum up to n. You are then left with [tex]
    \sum_{i=1}^{n+1} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)
    [/tex].

    Now you need to take (n+1)(n+2)(n+3) = 4[(n+1)(n+2)(n+3] / 4 and add the two fractions together. Then, yes, you simplify and find it is equal to what you desired.
     
  6. Feb 28, 2010 #5
    Did you take the (n+1)(n+2)(n+3) from the right side or from the i's? What happens to the extra (n+1)(n+2)(n+3) terms on the right?
     
  7. Feb 28, 2010 #6
    [tex]
    \sum_{i=1}^{n+1} i(i+1)(i+2) =
    (n+1)(n+2)(n+3) +
    \sum_{i=1}^{n} i(i+1)(i+2)
    [/tex]
    [tex]= (n+1)(n+2)(n+3)+
    \frac{n(n+1)(n+2)(n+3)}{4}
    [/tex] (by the induction hypothesis)


    [tex]
    = \frac{4(n+1)(n+2)(n+3)}{4} +
    \frac{n(n+1)(n+2)(n+3)}{4}

    [/tex]

    Add these fractions together and factor out (n+1)(n+2)(n+3)
     
  8. Feb 28, 2010 #7
    You completely lost me..

    What's on the left side and the right side?
     
  9. Feb 28, 2010 #8
    Ok so I simplified and it's

    1/4 (n+1) (n+2) (n+3) (n+4)

    So what do I do after I simplify?
     
  10. Feb 28, 2010 #9
    Can you show me where I lost you?
    You proved the base case. So that's done.
    Now the goal is to show that for the statement S(n), S(n) is true implies S(n+1) is true.
    So the goal in my last post was to take the sum to n+1 and try to reduce it to showing that it is equal to [(n+1)(n+2)(n+3)(n+4)]/4 (which is the n+1 case of the right hand right of the equality in your first post). In fact you had started this in your first post. You need to carry out all the simplification steps.
     
  11. Feb 28, 2010 #10
    Once you have achieved 1/4 (n+1) (n+2) (n+3) (n+4) after reducing from the original sum to n+1, you have finished the second step.
     
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