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Proof by induction

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove by induction $$w_k = w_{k−2} + k$$, for all integers $$k \ge 3, w_1 = 1,w_2 = 2$$ has an explicit formula
    $$ w_n =\begin{cases}
    \frac{(n+1)^2}{4}, & \text{if $n$ is odd} \\
    \frac n2(\frac n2 + 1), & \text{if $n$ is even}
    \end{cases}$$


    2. Relevant equations

    3. The attempt at a solution

    Inductive step for when n is odd:

    Suppose $$w_k = \frac{(k+1)^2}{4}$$ if k is odd. Then by definition of w, we have $$w_{k + 2} = w_k + k + 2 = \frac{(k+1)^2}{4} + k + 2 = \frac {k^2 + 2k + 1}{4} + k + 2= \frac {k^2 + 6k + 8}{4} = \frac {(k +3)^2}{4} $$ if k + 2 is odd.

    Is it important that we prove $$w_{k + 1} = \frac{(k+2)^2}{4}$$ if k+ 1 is odd or is the proof for $$w_{k + 2} = \frac{(k+3)^2}{4}$$ if k + 2 is enough?

    If none of the above makes sense, can I please get help getting started with the inductive step.
     
  2. jcsd
  3. Oct 13, 2016 #2

    PeroK

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    It's fine to here. That's the odd numbers sorted out. Now, what about the even numbers?
     
  4. Oct 13, 2016 #3

    fresh_42

    Staff: Mentor

    And the induction base? You have to verify this, too. It is really important and not just something annoying.
    And what if k is even?
    (Remark: there is a typo.)
     
  5. Oct 13, 2016 #4
    Thanks for the comments. Do we have to handle $$w_{k+1}$$ at all?
     
  6. Oct 13, 2016 #5

    PeroK

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    Why would you?
     
  7. Oct 13, 2016 #6
    I was worried we proved w_k for every other integer k. I see why that's wrong. Thanks.
     
  8. Oct 13, 2016 #7

    fresh_42

    Staff: Mentor

    With the induction base (two values), the odd numbers, which are covered by ##k \rightarrow k+2##, only the even are missing. What other numbers can you think of?
     
  9. Oct 13, 2016 #8
    Can't think of anything else :) Thanks.
     
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