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Help with mathematical assertions for natural numbers

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove by Mathematical Induction that the assertion,
    n
    ∑ r^2 = n/6 (n+1)(2n+1)
    r=1

    holds for every natural number n.


    2. Relevant equations

    Ok, so basically, how do you solve this question? I have got to the Induction step but I'm not sure how to do it.



    3. The attempt at a solution

    I've replaced n with k so I have,

    1^2 + 2^2 + 3^2 + ... k^2 = k/6 (k+1)(2k+1)

    Then I've added the (k+1)th term to each side to I have,

    1^2 + 2^2 + 3^2 + ... k^2 + (k+1)^2 = k/6 (k+1)(2k+1) + (k+1)^2

    So where do I go from here?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 10, 2011 #2

    I like Serena

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    Hi MegaDeth! :smile:

    Substitute (k+1) in n/6 (n+1)(2n+1) and check if that is equal to your expression.
     
  4. Nov 10, 2011 #3
    Sorry, but I'm not really sure how to do that :S How do I substitute it and equal it to the expression?
     
  5. Nov 10, 2011 #4
    To complete the proof, you have to show that the right side of the equation is (k+1)/6 ((k+1)+1)(2(k+1)+1). All it requires is algebra to re-express the right side in a way that shows your assertion is correct.
     
  6. Nov 10, 2011 #5

    I like Serena

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    If you substitute n=(k+1) in n/6 (n+1)(2n+1), you get:

    (k+1)/6 ((k+1)+1)(2(k+1)+1).


    Is that the same as k/6 (k+1)(2k+1) + (k+1)^2?
     
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