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Mathematical Induction - Problem 2
University of Michigan
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Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
We’re now going to use mathematical induction to prove that the sum of the first n odd integers is n². Using induction the first thing went to do is show that it works for n equals 1.
First thing we do, sum of the first terms 1 that's got to be the same thing as 1², okay that works.
Second thing is assume it works for a arbitrary value of k. So then we assume that 1 plus 3 plus 5 plus so on and so forth plus 2k minus 1 is equal to k². And then lastly we want to use the fact of this assumption to prove that it works for k plus 1. One term more. What we get in that case is 1 plus 3 plus 5 plus 2k minus 1 plus and then we have to plug in k plus 1 into our summation so this turns into 2(k plus 1 minus 1) and in theory this should equal quantity k plus 1².
Let’s see what we actually have done here. We have, this is just 2k plus 2 minus 1 which is then just 2k plus 1. What we have on the right side is, let’s use a different color, 1 plus 3 plus 5 plus dot, dot, dot, 2k minus 1 plus k plus 1. Using my assumption up here I already know that this whole thing is equal to k². What I really have on the right side is simply k² plus 2k plus 1. Checking what I have on the right side, if I were to FOIL this out I end up getting k² plus 2k plus 1.
We were able to prove that these two sides are equal therefore using our assumption that it works for k, we have proven that it works for k plus 1 as well.
Using mathematical induction, you show it works for the first thing, you assume it works for an arbitrary value k and then using that arbitrary value k you prove that it works for k plus 1.
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Sample Problems (6)
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Problem 1 6,813 views
1 + 2 + 3 +....... n = n(n + 1) 2
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Prove:1 + 3 + 5 +.....(2n - 1) = n²
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Problem 4 970 views
Problem 5 977 views
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