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anonymous

  • one year ago

Mathematical induction

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  1. anonymous
    • one year ago
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    Can someone help me with this? Don't just give me the answer but i need the steps too cuz i don't understand it. Thanks :)

  2. amistre64
    • one year ago
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    well we should start with the first step ... is it true for some value of n? are we restricted by our n values? these usually define a domain

  3. freckles
    • one year ago
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    that one line seems to be missing +2(k+1)+6 on the left hand side

  4. freckles
    • one year ago
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    wait I'm confused

  5. freckles
    • one year ago
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    it is also missing a square on the (k+1)

  6. amistre64
    • one year ago
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    lol, your giving out answers

  7. anonymous
    • one year ago
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    I'm confused

  8. freckles
    • one year ago
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    Me?

  9. amistre64
    • one year ago
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    what is the question? is it asking if the setup is correct?

  10. freckles
    • one year ago
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    I thought it was asking to use this to prove this

  11. freckles
    • one year ago
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    lol I didn't know I was giving answers

  12. anonymous
    • one year ago
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    yep it is asking if it is correct

  13. freckles
    • one year ago
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    oops

  14. amistre64
    • one year ago
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    then we would walk thru the process, to determine if the step they propose is valid.

  15. amistre64
    • one year ago
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    as freckles pointed out already ... its missing alot of stuff tho

  16. freckles
    • one year ago
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    I guess it kind of does make sense as a true false question.

  17. freckles
    • one year ago
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    hint: \[\text{ If give } P_n \text{ to find } P_{k+1} \text {just replace the } n \text{ 's with } (k+1)\]

  18. freckles
    • one year ago
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    on both left and right hand sides of the equation

  19. amistre64
    • one year ago
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    id say that replacement depends on the structure of the problem

  20. amistre64
    • one year ago
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    the proofing in this case requires us to add the (k+1)th term to each side and see if it simplifies to a format that looks as if all we did was replace n by (k+1)

  21. anonymous
    • one year ago
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    Oh wait so since when you substitute the n's for (k+1) the top equation has a k^2 and the bottom doesn't so it already wouldn't be equal?

  22. freckles
    • one year ago
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    right but there is also another thing missing on the left hand side

  23. anonymous
    • one year ago
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    would the left side turn into 2k+8 or something?

  24. freckles
    • one year ago
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    2(k+1)+6 2k+2+6 2k+8 yep that should be the last term on the left hand side where the one before it was 2k+6

  25. freckles
    • one year ago
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    you know because if it was 8+10+12+...+(2k+6) which is the exact thing as 8+10+12+...+(2n+6) except we have k instead of n then we should have this is k^2+7k since that other one is n^2+7n

  26. freckles
    • one year ago
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    And sorry about taking the fun out of the problem... I didn't realize the question was is this true or not at first.

  27. freckles
    • one year ago
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    I thought we were going to prove something. :( And there was just a lot of type-o.

  28. anonymous
    • one year ago
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    Its okay haha at least i understand it now. Thank you :)

  29. freckles
    • one year ago
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    and yeah the way we prove it is true for k+1 is take that previous thing replace n's with k's then as @amistre64 said had the (k+1)th term on both sides which you already know is +(2k+8) \[\color{red}{8+10+12+\cdots+(2k+6)}+(2k+8)=\color{red}{k^2+7k}+(2k+8) \\ \text{ where we want to show the right hand side actually does equal } \\ (k+1)^2+7(k+1)\]

  30. freckles
    • one year ago
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    and if you are doing that you might have an easier time expanding (k+1)^2+7(k+1) to show it is the same as k^2+7k+(2k+8)

  31. freckles
    • one year ago
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    but anyways this is getting into the actual proof of the P_n thing is true for all positive integer n

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