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mathmath333

  • one year ago

Fun question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \normalsize \text{Find the 20th term of the sequence }\hspace{.33em}\\~\\ &1,\ 5,\ 15,\ 34,\ 65\cdots\cdots \end{align}}\)

  2. ikram002p
    • one year ago
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    the magic lol

  3. ikram002p
    • one year ago
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    do u really know how does the magic squares works ? it would be fun to play with

  4. mathmath333
    • one year ago
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    idk magic squares

  5. ikram002p
    • one year ago
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    oh so u just know the formula that defines the suitable sum xD nvm

  6. mathmath333
    • one year ago
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    no i dont know any formula for this nth term

  7. ikram002p
    • one year ago
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    :) lets see if someone else know

  8. anonymous
    • one year ago
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    it s 4010 and here is the formula a(n) = n*(n^2 + 1)/2.

  9. mathmath333
    • one year ago
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    there may be the predefined nth term , but thats not any fun doing this

  10. anonymous
    • one year ago
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    @mathmath333 what do you think ?

  11. mathmath333
    • one year ago
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    i think we should do any observation

  12. anonymous
    • one year ago
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    so enjoy observing :)

  13. anonymous
    • one year ago
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    the first 20 term : 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010

  14. mathmath333
    • one year ago
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    how did u know the formula by the way

  15. ikram002p
    • one year ago
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    will ok i have this idea , we have 5 terms , so lets set a polynomial of order 4 \(\Large x_n= an^4+bn^3+cn^2+dn+e \\ \Large x_0=1 ~~thus e=1 \\ \Large x_1=a+b+c+d+1=5 \\ \Large x_2=16a+8b+4c+2d+1=15 \\\Large x_3=81a+27b+9c+3d+1=34\\ \Large x_4= 256a+64b+16c+4d+1=65 \) so now we have 4 equations with 4 variables :)

  16. mathmath333
    • one year ago
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    that looks little cumbersome

  17. anonymous
    • one year ago
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    you have to focus all the numbers and then doing some arbitrary tries to guess the formular . then you customize all the tries . and you get the right one . doing this is maybe innate or it needs a few of experience dealing with sequence

  18. ikram002p
    • one year ago
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    well this is one way to do it -.- u can make sine cos combination (fourier)

  19. ikram002p
    • one year ago
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    anyway this is wolfram answer for the new polynomial (as u dont wanna the predefined )

  20. anonymous
    • one year ago
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    @ikram002p gave you the normal right method to follow when dealing with those kind of sequence

  21. mathmath333
    • one year ago
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    ok a hint \(\large \color{black}{\begin{align} &1,\\~\\&\ 2+3,\\~\\&\ 4+5+6,\\~\\&\ 7+8+9+10,\\~\\&\ 11+12+13+14+15 \\~\\&\cdots\cdots \\~\\ \end{align}}\)

  22. ikram002p
    • one year ago
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    this is what been defined :O

  23. dan815
    • one year ago
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    oh simple

  24. ikram002p
    • one year ago
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    -.-

  25. dan815
    • one year ago
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    they just seem so close to doubling lol

  26. dan815
    • one year ago
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    |dw:1434620107872:dw|

  27. dan815
    • one year ago
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    |dw:1434620166190:dw|

  28. dan815
    • one year ago
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    |dw:1434620264189:dw|

  29. dan815
    • one year ago
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    solve a,b,c and u will ahve an equation that can get any xth term T(x)=...

  30. dan815
    • one year ago
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    I dont personally like this method though, we havent probven its a diff of 3 from there :)

  31. dan815
    • one year ago
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    but then again the equation u get does saitsfy all the terms uare given, so u cannot say its wrong

  32. dan815
    • one year ago
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    but then again the argument is that, there is obviously going to be some polynomial equation taht will will any given finite sequence,

  33. dan815
    • one year ago
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    as long as the polynomial equation solved is a sufficiently lower degree than the sequence given, or actually as long as its 1 degree lower than the number of tgerms given, we can say this polynomial equation solves the whole sequence

  34. mathmath333
    • one year ago
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    ok i will give a last hint

  35. dan815
    • one year ago
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    WAIT !! i solve it, u want me to do it another way!! dont tell me

  36. mathmath333
    • one year ago
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    ok

  37. dan815
    • one year ago
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    1 5 15 34 65 . . . 1 2+3 4+5+6 7+8+9+10

  38. dan815
    • one year ago
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    -.-

  39. dan815
    • one year ago
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    oaky so all we gotta do is calculate the nth term we will see at 20

  40. dan815
    • one year ago
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    and terms are going up at n*(n+1) so

  41. dan815
    • one year ago
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    divided by 2

  42. dan815
    • one year ago
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    have to know the sum of squares formula basically

  43. dan815
    • one year ago
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    |dw:1434620893662:dw|

  44. dan815
    • one year ago
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    |dw:1434620949098:dw|

  45. dan815
    • one year ago
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    oops wrong wait!!

  46. dan815
    • one year ago
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    i think i have ot subtract sum of series from that

  47. dan815
    • one year ago
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    |dw:1434621049221:dw|

  48. dan815
    • one year ago
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    |dw:1434621081983:dw|

  49. dan815
    • one year ago
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    is that the answer??

  50. dan815
    • one year ago
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    dont mind the scribbles before lol

  51. mathmath333
    • one year ago
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    wait lol

  52. dan815
    • one year ago
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    ok i go afk now

  53. dan815
    • one year ago
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    wait its add! omg

  54. dan815
    • one year ago
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    |dw:1434621259615:dw|

  55. mathmath333
    • one year ago
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    u have done it partially correct

  56. dan815
    • one year ago
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    maybe its subtract ugh xD

  57. dan815
    • one year ago
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    subtracting works i think

  58. dan815
    • one year ago
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    lemme just make it into a program i cant keep reyping that

  59. mathmath333
    • one year ago
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    well that works now this one , m still confuzed on how u got that but its correct \(\large \color{black}{\begin{align} k=\dfrac{n^2(n+1)}{2}-\dfrac{n(n-1)}{2 }\hspace{.33em}\\~\\ \end{align}}\)

  60. dan815
    • one year ago
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    :)

  61. dan815
    • one year ago
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    okay ill show u what i did its simple first i looked at this picture

  62. dan815
    • one year ago
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    |dw:1434621644190:dw|

  63. dan815
    • one year ago
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    we need to see the last term on the 20th line

  64. dan815
    • one year ago
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    and for every line we get 1 new term

  65. dan815
    • one year ago
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    so its the sum of increasing number by 1

  66. dan815
    • one year ago
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    um ya the main thing is to just find the nth term u are dealing with

  67. dan815
    • one year ago
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    to count the number of tertms, we see it has this form which is 1,2,3,4.... first term has 1 term,2nd term has 2 terms, 3rd term has 3 terms and so on... for 20th term we have 20 terms, and how many terms hae passed by till that time, that would be 1+2+3+4....+19

  68. dan815
    • one year ago
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    that is the number we are adding +1 +2+3... to +19

  69. dan815
    • one year ago
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    i feel like im failing at explaining this lol, it should be simple xD

  70. mathmath333
    • one year ago
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    the idea was to spot the triangular numbers at the last place that is similar to Dan815's idea \(\large \color{black}{\begin{align} \color{red}{1},\\~\\\ 2+\color{red}{3},\\~\\\ 4+5+\color{red}{6},\\~\\\ 7+8+9+\color{red}{10},\\~\\\ 11+12+13+14+\color{red}{15 }\\~\\\cdots\cdots \\~\\ \end{align}}\) and then the \(20th\) term would have 19 terms as the pattern shows so \(\large \color{black}{\begin{align} 191+192+\cdots+210 \\~\\ \end{align}}\) thats easy to add now

  71. dan815
    • one year ago
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    nice

  72. dan815
    • one year ago
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    thats a pretty different method from me

  73. dan815
    • one year ago
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    i did not notice that triangular number pattern there :O cool :)

  74. mathmath333
    • one year ago
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    i thought u did it that way

  75. dan815
    • one year ago
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    hehe no its a bit different

  76. dan815
    • one year ago
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    but its interting the triangular numbers came up anayway

  77. dan815
    • one year ago
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    interesting*

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