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Proof by induction

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data
    For n ∈ N (the natural numbers), ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2.

    2. Relevant equations

    For proof by induction, to show that the statement, P(n) is true for all n ∈ N you must show that P(1) is true and P(k+1) is true whenever P(k) is true.

    3. The attempt at a solution

    Ok so here is where I am at:
    We will use induction on n.
    Basis step:
    For n = 1, we have ∑ (-1)112 (sum from i=1 to 1) = (-1)112 = -1 and (-1)11(1+1)/2 = -1. So, when n =1, ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2.
    Inductive step:
    Suppose ∑ (-1)ii2 (sum from i=1 to k) = (-1)kk(k+1)/2. Then ∑ (-1)ii2 (sum from i=1 to k+1) = ∑ (-1)ii2 (sum from i=1 to k) + (-1)k+1(k+1) = (-1)kk(k+1)/2 + (-1)k+1(k+1) by the inductive hypothesis. So, -1kk(k+1)/2 + 2(-1)k+1(k+1)/2 = (-1kk(k+1) + 2(-1k+1)(k+1))/2 = ((-1k + -1k+1)(k+1)(k+1)+1)) /2.

    And here is where I am stuck. Cause right now I have the second part i.e. (k+1)((k+1)+1)/2 right, however I can't see how to get the exponents to become -1k+1

  2. jcsd
  3. Apr 20, 2010 #2


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    Your induction step should be looking at (-1)^k*k*(k+1)/2+(-1)^(k+1)*(k+1)^2. You seem to be missing the "^2" on the (k+1). Try it again. Factor out (-1)^(k+1) first.
    Last edited: Apr 20, 2010
  4. Apr 20, 2010 #3
    Dang. Now that I see that its so easy. Thanks.
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