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anonymous

  • one year ago

Prove the following proposition using the principle of mathematical induction

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  1. anonymous
    • one year ago
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  2. zepdrix
    • one year ago
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    \[\large\rm \sum_{i=1}^n (2i-1)=n^2,\qquad\forall n\in\mathbb Z^+\] Ok so we start with our base case: \(\large\rm n=1\)

  3. zepdrix
    • one year ago
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    \[\large\rm (2\cdot 1-1)=2-1=1\]Which is equal to \(\large\rm 1^2\), so our base case is satisfied, ya?

  4. anonymous
    • one year ago
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    yes

  5. zepdrix
    • one year ago
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    We'll assume it's true for \(\large\rm n=k\). Which means we assume \(\large\rm \color{orangered}{1+3+...+(2k-1)=k^2}\) is true.

  6. zepdrix
    • one year ago
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    We call this the Induction Hypothesis ^

  7. anonymous
    • one year ago
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    yeah

  8. anonymous
    • one year ago
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    okay

  9. zepdrix
    • one year ago
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    Then for our Induction Step: \(\large\rm n=k+1\) Let's see if we can get our formula set up correctly :d

  10. anonymous
    • one year ago
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    to equal n^2?

  11. zepdrix
    • one year ago
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    Well, notice that our last number in the sum will be \(\large\rm k+1\) So we won't end up with \(\large\rm n^2\), we'll end up with \(\large\rm (k+1)^2\) on the right. But setting up the left side is going to be a little tricky. Hopefully you can follow what I'm doing here.

  12. anonymous
    • one year ago
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    is it not k+2?

  13. zepdrix
    • one year ago
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    Your summation is giving you a bunch of terms like this:\[\large\rm (2\cdot\color{royalblue}{1}-1)+(2\cdot\color{royalblue}{2}-1)+...+(2\color{royalblue}{k}-1)+(2\color{royalblue}{(k+1)}-1)=(k+1)^2\]

  14. zepdrix
    • one year ago
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    I'll simplify the first few terms like I did the last time.\[\large\rm 1+3+...+(2k-1)+(2(k+1)-1)=(k+1)^2\]

  15. zepdrix
    • one year ago
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    k+2? what? :o

  16. anonymous
    • one year ago
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    i dont get the part (2(k+1)−1)

  17. anonymous
    • one year ago
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    i dont get where k+1 came from

  18. anonymous
    • one year ago
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    wait i get it

  19. anonymous
    • one year ago
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    i get it

  20. zepdrix
    • one year ago
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    XD

  21. zepdrix
    • one year ago
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    So in the Induction Hypothesis, we let \(\large\rm k\) be the largest number in the sequence. We replaced n by k. In our Induction Step, we're letting \(\large\rm k+1\) be the largest number in the sequence. It's the number that comes after k. So our summation grows a little bit, by one more term. And it should be equivalent to (k+1)^2 now, instead of k^2.

  22. anonymous
    • one year ago
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    yup

  23. zepdrix
    • one year ago
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    If you scroll up, you'll notice I colored our Induction Hypothesis in orange. We need to relate this back to that orange thing.

  24. anonymous
    • one year ago
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    1+3+...+(2k−1)=k^2

  25. anonymous
    • one year ago
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    this?

  26. zepdrix
    • one year ago
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    Yes, that entire left side is contained within our Induction Step formula.

  27. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{1+3+...+(2k-1)}+(2(k+1)-1)=(k+1)^2\]Do you see it? :)

  28. anonymous
    • one year ago
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    yeah

  29. zepdrix
    • one year ago
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    Since our Induction Hypothesis tells us that the orange stuff is k^2, we can make the switch,\[\large\rm \color{orangered}{k^2}+(2(k+1)-1)=(k+1)^2\]

  30. anonymous
    • one year ago
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    ohhhhh oh kay

  31. zepdrix
    • one year ago
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    From there, you need to turn the left side into (k+1)^2 somehow. Do some Algebra steps expand out the stuff, then try to factor it down.

  32. anonymous
    • one year ago
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    thank yyou!

  33. zepdrix
    • one year ago
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    np \c:/

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