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anonymous

  • one year ago

A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work. Sn: 12 + 42 + 72 + . . . + (3n - 2)2 = 𝑛(6𝑛2−3𝑛−1)2

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  1. anonymous
    • one year ago
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    n(6n^2-3n-1)/2

  2. anonymous
    • one year ago
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    anybody got a clue

  3. anonymous
    • one year ago
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    @anonymous_user

  4. phi
    • one year ago
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    can you make this clearer (3n - 2)2 ?

  5. anonymous
    • one year ago
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    n(6n^2-3n-1)/2

  6. phi
    • one year ago
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    No, I mean the last term in the sum 12 + 42 + 72 + . . . + (3n - 2)2

  7. anonymous
    • one year ago
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    its 1^2+4^2+7^2

  8. anonymous
    • one year ago
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    im so lost

  9. phi
    • one year ago
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    oh. it looks like twelve + forty-two+... and I was not making any progress (who would have guessed)

  10. anonymous
    • one year ago
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    lol its my fualt i should have double checked thst sooo

  11. phi
    • one year ago
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    Do you have to prove the result or derive it? proving it should be easier

  12. anonymous
    • one year ago
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    no i just have to show thast the stamnets are true and show all my work proving them

  13. phi
    • one year ago
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    are you studying proof by induction?

  14. anonymous
    • one year ago
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    i think so

  15. anonymous
    • one year ago
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    yes iam

  16. phi
    • one year ago
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    the first step is show the identity is true for n=1 can you do that ?

  17. anonymous
    • one year ago
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    no im jsut now learning this stug=ff lol sorry im a newbie

  18. phi
    • one year ago
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    when n=1, what is "series" ?

  19. phi
    • one year ago
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    what is the last term of the summation when n=1 ?

  20. anonymous
    • one year ago
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    1

  21. phi
    • one year ago
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    and the starting term is 1 in other words the entire summation is 1^2 (just 1 term) on the right-hand side we have the formula n(6n^2-3n-1)/2 what is that when n=1 ?

  22. anonymous
    • one year ago
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    ummmmm

  23. phi
    • one year ago
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    you sub in n=1 into the formula n(6n^2-3n-1)/2 what do you get ?

  24. anonymous
    • one year ago
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    scratches head um n+1(6n^2-3n-1)/2

  25. phi
    • one year ago
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    the idea is everywhere you see n in the formula, erase it, and write in 1 in its place then simplify (now that you have a number instead of a letter, you can do that) what do you get ?

  26. anonymous
    • one year ago
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    1(61^2-31-1)/2

  27. phi
    • one year ago
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    ok. and remember when you have for example 3n that means 3*n ( multiply signs are usually left out) so the formula is 1*(6*1^2 - 3*1 -1)/2 now simplify that

  28. anonymous
    • one year ago
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    hold on im doing it now

  29. anonymous
    • one year ago
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    the answer is 1 right

  30. phi
    • one year ago
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    yes, it simplifies to (6-4)/2 = 2/2 = 1 that shows the formula works for n=1 because 1^2 = 1 (the answer we get from the formula)

  31. phi
    • one year ago
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    now the hard part. we assume the formula works for numbers up to n (whatever n happens to be) and show it also works for n+1 This part is difficult because it looks like we need some messy algebra.

  32. anonymous
    • one year ago
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    this is what i didi 1^2 evaluates to 1 Multiply 1 and 6 1 6*1^2 evaluates to 6 Multiply 1 and 3 1 3*1 evaluates to 3 6*1^2-3*1 evaluates to 3 6*1^2-3*1-1 evaluates to 2 Multiply 1 and 2 1 1*(6*1^2-3*1-1) evaluates to 2 1*(6*1^2-3*1-1)/2 evaluates to 1

  33. anonymous
    • one year ago
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    uhhh math

  34. anonymous
    • one year ago
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    so how dowe start this

  35. phi
    • one year ago
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    yes, that is how you evaluate it.

  36. anonymous
    • one year ago
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    sweet

  37. phi
    • one year ago
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    all of that proves that the formula works for n=1 now assume the formula works for numbers up to n show it works for n+1 Induction Step: Assume the identity is true for n: \[ 1^2 + 4^2 + ... +(3n-1)^2 = \frac{n(6n^2-3n-1)}{2} \]

  38. phi
    • one year ago
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    now add the next term to the summation that is add \( \left(3(n+1)-1 \right)^2 \) to both sides \[ 1^2 + 4^2 + ... +(3n-1)^2+\left(3(n+1)-1 \right)^2 = \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2\]

  39. anonymous
    • one year ago
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    holy moly tat is hard

  40. phi
    • one year ago
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    yes, it is very difficult.

  41. anonymous
    • one year ago
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    so is there more too it plz say no lol

  42. phi
    • one year ago
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    the formula for n+1 becomes \[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \] we have to show that \[ \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2 \] can be written as \[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \]

  43. anonymous
    • one year ago
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    thnx

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