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Am I on the right track with this induction question?

  1. Apr 26, 2004 #1
    Step 0)
    Code (Text):

    2[sup]2[/sup] + 4[sup]2[/sup] + 6[sup]2[/sup]+...+(2n)[sup]2[/sup] = 2n(n+1)(2n+1)/3

    Step 1)
    Let n = 2
    2[sup]2[/sup] + 4[sup]2[/sup]
    = 4 + 16
    =  20

    2(2)(2+1)(2(2)+1)/3
    = 60/3
    = 20

    Step2)
    Assume that the formula works for n=1,2,3,...,k
    ie. 2[sup]2[/sup] + 4[sup]2[/sup] + 6[sup]2[/sup]+...+(2n)[sup]2[/sup] = 2k(k+1)(2k+1)/3
    Step 3)
    2n(n+1)(2n+1)/3 + (k+1)[sup]2[/sup]

     
    ?
     
    Last edited: Apr 26, 2004
  2. jcsd
  3. Apr 26, 2004 #2

    arildno

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    Science Advisor
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    Dearly Missed

    Step 1) is a correct verification of the formula.
    Step 2) is very confusing, you seem to use n and k interchangeably!
    n is the summation index, while the k'th term is the last to be summed!
    Now instead:
    Let 1<=n <= k:
    Assume the proposition holds for the choice k.
    We are to show that the proposition holds when summing 1<=n<=k+1:

    Sum(from 1 to k+1)=Sum(from 1 to k)+(2(k+1))^(2)=

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

    Now rearrange and try to gain the "formulaic prediction" for k+1.
     
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