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Bee_see

  • one year ago

Prove by induction that 12 + 22 + · · · + n2 = n(n + 1)(2n + 1)/6 for every positive integer n.

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  1. amistre64
    • one year ago
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    what should our first step be?

  2. Bee_see
    • one year ago
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    I have written some stuff for the problem...so..|dw:1441663294530:dw|

  3. Bee_see
    • one year ago
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    assuming Sk is true

  4. amistre64
    • one year ago
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    your #2 is a little wrong,|dw:1441663385737:dw|

  5. Bee_see
    • one year ago
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    |dw:1441663404145:dw|

  6. amistre64
    • one year ago
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    i tend to use P(k) \[P(k)=12+22+...+k^2=\frac{k(k+1)(2k+1)}{6}\] or simply \[P(k)=\frac{k(k+1)(2k+1)}{6}\] now the idea is to add the next term to each side ... what is the next term?

  7. welshfella
    • one year ago
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    now add (k+ 1)^2 to the sum of k terms

  8. Bee_see
    • one year ago
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    (k+1)^2

  9. Bee_see
    • one year ago
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    |dw:1441663539407:dw|

  10. amistre64
    • one year ago
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    \[P(k)+(k+1)^2~\color{red}{?=?}~\frac{k(k+1)(2k+1)}{6}+(k+1)^2\] the idea is to get it into the same FORM, since the only thing we really know is that the FORM is good for k=1

  11. Bee_see
    • one year ago
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    so k^2? before (k+1)^2 on the left side?

  12. Bee_see
    • one year ago
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    no K^2*

  13. amistre64
    • one year ago
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    P(k) is equal to all the kth term stuff, so all we need to do is add the next term to it can we rework thru valid mathimatical procesess \[\frac{k(k+1)(2k+1)}{6}+(k+1)^2\color{red}{\implies}\frac{(k+1)^2((k+1)+1)(2(k+1)+1)}{6}\]

  14. amistre64
    • one year ago
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    bah, i put a ^2 where there shouldnt be one ...

  15. amistre64
    • one year ago
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    \[\frac{k(k+1)(2k+1)}{6}+(k+1)^2\color{red}{\implies}\frac{(k+1)~((k+1)+1)~(2(k+1)+1)}{6}\]

  16. amistre64
    • one year ago
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    as long as we can work it thru from left format, to the right format, we are good as far as proof by induction goes

  17. Bee_see
    • one year ago
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    why didn't you do k(k+1) and turned it into k^2+k?

  18. amistre64
    • one year ago
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    because that is NOT the format that we are trying to prove ...

  19. amistre64
    • one year ago
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    proof by induction is ALL about the FORMAT ... we cannot go changing the format

  20. amistre64
    • one year ago
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    we want to work the (k+1)th term INTO the format ... not the other way around.

  21. Bee_see
    • one year ago
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    how did you get rid of the (k+1)^2? on the right side?

  22. amistre64
    • one year ago
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    i didnt ... i just simply showed that we need to go from one form, to the other ...

  23. amistre64
    • one year ago
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    id start by getting it all over the same denominator, then parsing out the numerator

  24. amistre64
    • one year ago
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    now, as a guide, you can work from the left format into the right format ... but that only proves you can work it one way, and not back again ... as long as you can show valid mathematical processes to work it back again your proof is done

  25. amistre64
    • one year ago
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    err, right into left ... never was good with sides lol

  26. Bee_see
    • one year ago
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    doesn't the numerator become k(k+1)(2k+1)+6(k+1)^2 ?

  27. amistre64
    • one year ago
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    yes

  28. amistre64
    • one year ago
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    now we can factor our a (k+1) to get started

  29. amistre64
    • one year ago
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    k(k+1)(2k+1)+6(k+1)^2 ^^^ ^^^ (k+1) [(2k+1)+6(k+1)] since we want to start it with a (k+1) we have that covered

  30. Bee_see
    • one year ago
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    ok

  31. Bee_see
    • one year ago
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    that still doesn't get rid of that 6 on the numerator

  32. amistre64
    • one year ago
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    :) lets guide ourselves backwards we want to end up with ((k+1)+1)(2(k+1)+1) (k+2)(2k+2+1) (k+2)(2k+3) 2k^2 +4k +3k +6 2k^2 +7k +6 ..................... expand what we got k(2k+1)+6(k+1) 2k^2 +k +6k +6 2k^2 +7k +6 they match up, so lets just work the process backwards

  33. amistre64
    • one year ago
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    now that we have a path to work from ... we can expand everything to here: 2k^2 +7k +6 split the 7k into 3k + 4k 2k^2 +3k + 4k +6 and facotr

  34. Bee_see
    • one year ago
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    I thought we wanted to end up with (k+1)[ (k+1)+1) ] [2(k+1)+1]...aren't you missing the (k+1)?

  35. amistre64
    • one year ago
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    no, im not missing it, im just ignoring it at the moment so i dont have to keep writing it down when we no longer have to focus on it ....

  36. Bee_see
    • one year ago
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    ok

  37. amistre64
    • one year ago
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    we already factored out a (k+1) ... no need to worry about it.

  38. amistre64
    • one year ago
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    \[\frac{(k+1)\color{red}{[k(2k+1)+6(k+1)]}}{6}\]

  39. amistre64
    • one year ago
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    \[\frac{`\color{red}{[k(2k+1)+6(k+1)]}}{`}\] \[\frac{`\color{red}{[2k^2+7k+6]}}{`}\] \[\frac{`\color{red}{[2k^2+\color{green}{(4k+3k)}+6]}}{`}\] \[\frac{`\color{red}{[(2k^2+4k)+(3k+6)]}}{`}\] \[\frac{`\color{red}{[2k(k+2)+3(k+2)]}}{`}\] \[\frac{`\color{red}{(k+2)(2k+3)}}{`}\] \[\frac{`\color{red}{(k+2)(2k+\color{green}{(2+1)})}}{`}\] \[\frac{`\color{red}{(k+2)((2k+2)+1)}}{`}\] etc ...

  40. Bee_see
    • one year ago
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    ok

  41. Bee_see
    • one year ago
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    I get it now.

  42. amistre64
    • one year ago
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    good luck then :)

  43. Bee_see
    • one year ago
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    Thanks for the help.

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