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Introductory mathematical induction problem

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data
    I am just learning the joys of mathematical induction, and this problem is giving me fits.


    2. Relevant equations
    I am trying to prove that 2 + 4 + 6 + … + 2n = [2n(n+1)]/2


    3. The attempt at a solution
    The base case is to prove P(1) is correct. Simple enough -- 2 = [2 x 1 (1+1)]/2. The RHS does equal 2, so we are good to go there.

    Next, substitute k for n. So, now we have 2 + 4 + 6 + … + 2k = [2k(k+1)]/2

    We have to replace k with (k + 1), which is what we need to prove in this proof. We can also rewrite the LHS to show 2k + 2(k+1) = [2k(k+1)]/2 -- I suspect this is the step where I have gone off the rails.

    With a little bit of algebra, I should be able to multiply the equations pout and prove the LHS = RHS, but I am somehow missing something, and I am pretty sure it is in step 3. The example problems I have followed in my text show the algebra is similar problems being pretty straightforward, but I am just not getting how to rewrite the substitution for k with (k + 1) as something mathematically correct.

    Thank you for your help in advance.

     
    Last edited by a moderator: Mar 27, 2015
  2. jcsd
  3. Mar 26, 2015 #2

    SammyS

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    Right. That's what you assume is true for some k where k ≥ 1.

    With the above assumption, you now have to prove that your statement is true for k + 1 .

    In other words, (with the above assumption) you need to prove that 2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2 .
     
    Last edited by a moderator: Mar 27, 2015
  4. Mar 26, 2015 #3
    OK, we can assume all of the constants to be true, and remove them form the proof. So, I now have

    2k + 2(k+1) = [2(k + 1)(k +1 +1)]/2

    So,

    2k + 2(k + 1) = (k + 1) (k + 2)
    2
    So, if I assume the constants to be true and can drop them from the proof, I have 2k + 2(k +1) = [2(k + 1)(k +1 +1)]/2

    So,

    2k + 2(k +1) = [2(k + 1)(k +1 +1)]/2
    2k + 2k + 2 = (k + 1) (k + 2)
    4k + 2 = k^2 + 3k + 2

    And here is the step that has been derailing me, in some way, shape, or form, since I started working on this thing -- what am I doing wrong here?
     
  5. Mar 26, 2015 #4

    SammyS

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    First of all, what are you calling a "constant" ?

    You can't assume that the following is true!
    2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2​
    You have to show (prove) that it's true assuming that the following is true. You have to show that it's a logical consequence of the following being true.
    2 + 4 + 6 + … + 2k = [2k(k+1)]/2​

    One plan of attack is to start with 2 + 4 + 6 + … + 2k = [2k(k+1)]/2 and then add 2(k+1) to both sides.
     
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