1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook