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Lukecrayonz

  • 4 years ago

Use sigma notation to write the sum: [1-(1/6)^2]+[1-(2/6)^2]+...1-(6/6)^2]

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  1. Lukecrayonz
    • 4 years ago
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    Im not 100% sure, but i'm thinking its this?

  2. Lukecrayonz
    • 4 years ago
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    |dw:1330750567546:dw|

  3. AccessDenied
    • 4 years ago
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    looks correct to me

  4. Hunus
    • 4 years ago
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    That's correct.

  5. mathg8
    • 4 years ago
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    good work

  6. AccessDenied
    • 4 years ago
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    with the squared, since im really good at not noticing that

  7. AccessDenied
    • 4 years ago
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    \[ \sum_{n=1}^{6} (1-(\frac{n}{6})^{2}) = (1-(\frac{1}{6})^{2}) + (1-(\frac{2}{6})^{2}) + ... + (1-(\frac{6}{6})^2) \] :D

  8. Lukecrayonz
    • 4 years ago
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    I'm trying to do a whole chapter in one night, so let's pull an all nighter?!:D

  9. Lukecrayonz
    • 4 years ago
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    Uhh, im stuck D:

  10. Lukecrayonz
    • 4 years ago
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    Write the first five terms of the arithmetic sequence. Find the common difference and write the nth term of the sequence as a function of n.

  11. AccessDenied
    • 4 years ago
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    i came online to find method of approximating square roots, snce i didnt want to pull out calculator. lol

  12. Lukecrayonz
    • 4 years ago
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    a[1]=0.375, a[k+1]=a[k]+0.25

  13. AccessDenied
    • 4 years ago
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    seems like the common difference would be 0.25 since k and k+1 are successive terms of the sequence and we add 0.25 to a(k) to get a(k+1), i gotta check that again to make sure -- it's been a while since i did this stuff. ;x

  14. Lukecrayonz
    • 4 years ago
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    Here's an example answer, if this helps. http://screensnapr.com/v/2PVSpz.png

  15. AccessDenied
    • 4 years ago
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    a_(k+1) = a_1 + (k+1 -1)d = a_1 + kd a_k = a_1 + (k-1)d = a_1 + kd - d = (a_(k+1)) - d --> a_(k+1) = a_1 + kd - d + d = (a_k) + d yeah, sounds about right 0.25 being the common difference, i'd just add 0.25 to the first term five times 0.375 + 0.25 = 0.625 0.625 + 0.25 = 0.845 etc

  16. Lukecrayonz
    • 4 years ago
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    I THINK it's a[2]=0.375+0.25=.625 a[3]=.625+0.25=0.875

  17. AccessDenied
    • 4 years ago
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    oh, i added wrong, 0.875 -not using calculator-

  18. Lukecrayonz
    • 4 years ago
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    a[4]0.875+.25=1.125 a[5]1.125+.25=1.375

  19. Lukecrayonz
    • 4 years ago
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    When you do a_k, do you mean sub k? Because when I put sub I put [k]

  20. AccessDenied
    • 4 years ago
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    yeah, when you use equation thing, sub-k is a_k, so that's why I use the _'s

  21. Lukecrayonz
    • 4 years ago
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    So when I'm doing the nth term of the sequence and writing the function for it, what would that be?

  22. Lukecrayonz
    • 4 years ago
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    a[n].375n+.25?

  23. AccessDenied
    • 4 years ago
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    A(n) = a_1 + (n-1)d; d = common difference, a_1 = first term would be the general form I'd use

  24. Lukecrayonz
    • 4 years ago
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    Thank you! I think mine would work but yours seems more sophisticated.

  25. Lukecrayonz
    • 4 years ago
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    Oh hey, your formula is the exact formula I'm using for my next questions.

  26. Lukecrayonz
    • 4 years ago
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    a[1]=0, d=-2/3

  27. AccessDenied
    • 4 years ago
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    well, if you test n=1, .375 + .25 = .625, which was the second term

  28. Lukecrayonz
    • 4 years ago
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    0+(n-1)(-2/3)

  29. AccessDenied
    • 4 years ago
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    yep, looks correct there. :D I don't know of a more intuitive way to remember the formula except just using it often, since I actually forgot it since going over it in Algebra, though.

  30. Lukecrayonz
    • 4 years ago
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    hmm, another question..

  31. Lukecrayonz
    • 4 years ago
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    Find the nth partial sum of the arithmetic series, a[1]=1.5 d=.5 n=20 If it matters, the series is 1.5, 1.45, 1.40, 1.35

  32. Lukecrayonz
    • 4 years ago
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    I think the formula S[n]=n/2(a[1]+a[n]) would be used?

  33. Lukecrayonz
    • 4 years ago
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    Just kidding far from that.

  34. Lukecrayonz
    • 4 years ago
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    Don't even have a a[n]! Haha, what was I thinking :P

  35. AccessDenied
    • 4 years ago
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    my source with the formulas for this says the (a[1] + a[n] is in numerator

  36. Lukecrayonz
    • 4 years ago
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    Here's an example answer, maybe this could clear up what the formula is..? http://screensnapr.com/v/zVwwkZ.png

  37. AccessDenied
    • 4 years ago
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    yeah, that looks accurate to the formula I found \[ S_n = \frac{n(a_1 + a_n)}{2} \] I guess it wants you to find the arithmetic series equation, substitute in n=20 to it, and then use the formula for the nth partial sum

  38. Lukecrayonz
    • 4 years ago
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    http://screensnapr.com/v/x2890d.png But in that, where the hell did 9 come from?

  39. Lukecrayonz
    • 4 years ago
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    n-1?

  40. AccessDenied
    • 4 years ago
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    n-1, n = 10 => 10-1 = 9

  41. Lukecrayonz
    • 4 years ago
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    So i got a[10]=1.5+19(.5)=11

  42. Lukecrayonz
    • 4 years ago
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    a[20] sorry

  43. AccessDenied
    • 4 years ago
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    yep, that's what I got as well

  44. Lukecrayonz
    • 4 years ago
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    Then i got s[20]=10(1.5+11)

  45. Lukecrayonz
    • 4 years ago
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    which equals 125

  46. AccessDenied
    • 4 years ago
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    Yep, got 125 as well. :)

  47. Lukecrayonz
    • 4 years ago
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    BOO! REAL WORLD APPLICATION PROBLEMS!

  48. Lukecrayonz
    • 4 years ago
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    http://screensnapr.com/v/YRvm1X.png

  49. Lukecrayonz
    • 4 years ago
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    15*16*17*18*19*20*21? Haha

  50. Lukecrayonz
    • 4 years ago
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    Adding not multiplying, that would be an incredible amount of logs.

  51. Lukecrayonz
    • 4 years ago
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    Uhh, a[1]=15, a[6]=21 and s[6]=21/2(15+21)=126

  52. AccessDenied
    • 4 years ago
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    i think the seventh term should be 21 since there are seven rows, with the first row being 15, sevond is 16, third is 17, fourth = 18, 5th = 19, 6th = 20, and 7th = 21

  53. AccessDenied
    • 4 years ago
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    but it does appear that 126 is correct.

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