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sparky16

  • 2 years ago

Can someone help with this? :) It is posted below in the comment section so I can use the Equation editor.

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  1. sparky16
    • 2 years ago
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    |dw:1362285707700:dw|\[\Sigma \]

  2. sparky16
    • 2 years ago
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    Find the sum of the arithmetic series! Pleassse help:)

  3. hartnn
    • 2 years ago
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    \(\huge \sum \limits_{k=1}^nk = \dfrac{n(n+1)}{2}\) and a constant (here, 5) can be taken out of sum sign... :)

  4. sparky16
    • 2 years ago
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    Where r we getting this formula? I'm confused where to start

  5. hartnn
    • 2 years ago
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    thats a standard formula that can be used.

  6. hartnn
    • 2 years ago
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    do you want a formula for sum of terms in arithmetic series ? its equivalent to this formula

  7. terenzreignz
    • 2 years ago
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    Deriving the formula for an arithmetic series?

  8. sparky16
    • 2 years ago
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    Yes that is the formula for the arithmetic series. Where do we go from there?:)

  9. sparky16
    • 2 years ago
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    to find the sum?

  10. terenzreignz
    • 2 years ago
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    Well, maybe best to use this formula instead \[\huge \sum \limits_{k=1}^nk = \dfrac{n(n+1)}{2}\] It's simpler :D

  11. terenzreignz
    • 2 years ago
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    So anyway, this sum is just... |dw:1362286375802:dw| right?

  12. hartnn
    • 2 years ago
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    ok, i'll mention the sum of arithmetic series formula also, then you choose which one you wanna use, the series will be 5,10,15,20,.... right ? (with n= 100 terms) he 1st term a1 = 5 and there's a common difference of d=5 then the sum formula is : \(S_n = (n/2)(2a+(n-1)d)\)

  13. sparky16
    • 2 years ago
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    terenzreignz...how do you get these expanded form answer? Can u explain that? sorry

  14. sparky16
    • 2 years ago
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    It is isn't the right answer in the back of the book

  15. terenzreignz
    • 2 years ago
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    That's just kind of the definition. Or you could say the sigma notation is a compact form of the sum. \[\huge \sum_{k=1}^n=1+2+3+...+(n-1)+n\] Basically adding all integers starting at 1 to n.

  16. terenzreignz
    • 2 years ago
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    \[\huge \sum_{k=1}^nk=1+2+3+...+(n-1)+n\] Bloody typo, sorry :)

  17. terenzreignz
    • 2 years ago
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    So anyway, we can rearrange...|dw:1362286994735:dw|

  18. terenzreignz
    • 2 years ago
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    And notice that these terms...|dw:1362287037927:dw| When added, they all equal n+1

  19. sparky16
    • 2 years ago
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    what is the final answer? the answer is supposed to be 25,250...I just don't know how to get it

  20. hartnn
    • 2 years ago
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    did you try any one of the 2 formula i gave u ??

  21. sparky16
    • 2 years ago
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    Going back to hartnn's response near the beginning. This is like the only formua I'm understanding how to do. \[S _{100}= (100(5 + ?))/2\] I'm just confused how to get that number I didn't fill in? It's supposed to be \[t _{n}\] (which is last term)

  22. sparky16
    • 2 years ago
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    Anyone?

  23. hartnn
    • 2 years ago
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    oh, you need last term ? the general term is 5k to get last term, just put k=100 in general term so, what u get as last term ?

  24. sparky16
    • 2 years ago
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    So i fill in that question mark spot with 100? Im not sure

  25. hartnn
    • 2 years ago
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    general term =5k with k=100, last term = 5*100 = 500 -_-

  26. sparky16
    • 2 years ago
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    To clarify, this is the formula I am using. \[S _{n}=\frac{ n(t _{1}+ t _{n} )}{ 2 }\]

  27. sparky16
    • 2 years ago
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    wait how does k =100 again?

  28. hartnn
    • 2 years ago
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    yup, thats the only other i didn't mention :P because you are summing from k=1 (which is 1st term) to k=100 (WHICH IS LAST TERM)

  29. hartnn
    • 2 years ago
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    lower limit givs first term, upper limit gives last term

  30. sparky16
    • 2 years ago
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    ok so 100 is the last term... And why do we multiply it by 5?

  31. hartnn
    • 2 years ago
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    last value of k is 100 general term = 5k so last term = 5*100

  32. sparky16
    • 2 years ago
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    and the difference between the numbers would be five right? or is it something else

  33. hartnn
    • 2 years ago
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    yes, common difference = 2nd term -1st term = 3rd term -2nd term =.... = 10-5 = 15-10 =.... = 5 but thats not required for the formula you are using

  34. sparky16
    • 2 years ago
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    Oh ok that makes sense now. Maybe you should help me on other problems I post. Your a good help!

  35. hartnn
    • 2 years ago
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    ok, sure, but did u get 25250 for this one ?

  36. sparky16
    • 2 years ago
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    Oh wait quick question! What was \[t _{1}\] in this equation

  37. hartnn
    • 2 years ago
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    t1 is the 1st term index represents which term

  38. hartnn
    • 2 years ago
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    tn is the n'th term

  39. sparky16
    • 2 years ago
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    the index is which part again?

  40. hartnn
    • 2 years ago
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    index is written as subscript \(\huge t_n\) here n is index and represent which term, t1 = 1st term, t2 =2nd term, and so on

  41. sparky16
    • 2 years ago
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    so t1 is 5

  42. hartnn
    • 2 years ago
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    yup. as 5 is the 1st term

  43. hartnn
    • 2 years ago
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    any more doubts ?

  44. sparky16
    • 2 years ago
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    ok im sorry so I'm trying to solve it on paper. And I'm confused again.. how did we get \[t _{n}\] in the formula

  45. sparky16
    • 2 years ago
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    oh wait my bad! i got it

  46. hartnn
    • 2 years ago
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    you mean in \(S_n = n (t_1+t_n)/2\) thats a general formula, do you want a derivation of this ?

  47. hartnn
    • 2 years ago
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    oh u got it ? good :)

  48. sparky16
    • 2 years ago
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    What would happen when I solve for \[t _{n}\] in an equation where n did not equal 1 and it equaled something else?

  49. hartnn
    • 2 years ago
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    didn't get your question ? do u have specific example for that ? say, if you want to find \(t_5\), then u put n=5 in general term tn

  50. sparky16
    • 2 years ago
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    |dw:1362289579021:dw| Like this. N does not equal 1

  51. hartnn
    • 2 years ago
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    i think u have made up this question ? or is it from your book/notes ?

  52. sparky16
    • 2 years ago
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    I made it up

  53. sparky16
    • 2 years ago
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    here Ill give you one from the book. |dw:1362289721775:dw|

  54. sparky16
    • 2 years ago
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    thats 30- m by the way. It kinda got cut off

  55. hartnn
    • 2 years ago
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    so, here your 1st term will start from m=10 in 30-m and last term you'll get by putting m=20 in 30-m

  56. sparky16
    • 2 years ago
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    so how would i get \[t _{n}\]

  57. hartnn
    • 2 years ago
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    tm = last term, last m =30 so to get last term tm, put m=30 in general term 30-m is this confusing ?

  58. sparky16
    • 2 years ago
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    kinda

  59. hartnn
    • 2 years ago
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    general term tm =30-m fist term t10 = 30-10 (put m=10) last term t20 = 30-20 (put m=20)

  60. sparky16
    • 2 years ago
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    so I would do 30 -10 first , and when do I do the 20? i dont know if that makes sense

  61. hartnn
    • 2 years ago
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    it doesn't -_-

  62. sparky16
    • 2 years ago
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    ok lets not do this problem.. Ill ask my teacher on monday..

  63. hartnn
    • 2 years ago
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    you have formula, you have n, you have 1st term, u have last term....just plug in values! ohh..ok, as u wish

  64. sparky16
    • 2 years ago
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    Ill be posting other problems so u may help me if u want

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