## mathmath333 one year ago Fun question

1. mathmath333

\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

the magic lol

3. ikram002p

do u really know how does the magic squares works ? it would be fun to play with

4. mathmath333

idk magic squares

5. ikram002p

oh so u just know the formula that defines the suitable sum xD nvm

6. mathmath333

no i dont know any formula for this nth term

7. ikram002p

:) lets see if someone else know

8. anonymous

it s 4010 and here is the formula a(n) = n*(n^2 + 1)/2.

9. mathmath333

there may be the predefined nth term , but thats not any fun doing this

10. anonymous

@mathmath333 what do you think ?

11. mathmath333

i think we should do any observation

12. anonymous

so enjoy observing :)

13. anonymous

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

how did u know the formula by the way

15. ikram002p

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

that looks little cumbersome

17. anonymous

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

well this is one way to do it -.- u can make sine cos combination (fourier)

19. ikram002p

anyway this is wolfram answer for the new polynomial (as u dont wanna the predefined )

20. anonymous

@ikram002p gave you the normal right method to follow when dealing with those kind of sequence

21. ikram002p
22. mathmath333

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}}

23. ikram002p

this is what been defined :O

24. dan815

oh simple

25. ikram002p

-.-

26. dan815

they just seem so close to doubling lol

27. dan815

|dw:1434620107872:dw|

28. dan815

|dw:1434620166190:dw|

29. dan815

|dw:1434620264189:dw|

30. dan815

solve a,b,c and u will ahve an equation that can get any xth term T(x)=...

31. dan815

I dont personally like this method though, we havent probven its a diff of 3 from there :)

32. dan815

but then again the equation u get does saitsfy all the terms uare given, so u cannot say its wrong

33. dan815

but then again the argument is that, there is obviously going to be some polynomial equation taht will will any given finite sequence,

34. dan815

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

35. mathmath333

ok i will give a last hint

36. dan815

WAIT !! i solve it, u want me to do it another way!! dont tell me

37. mathmath333

ok

38. dan815

1 5 15 34 65 . . . 1 2+3 4+5+6 7+8+9+10

39. dan815

-.-

40. dan815

oaky so all we gotta do is calculate the nth term we will see at 20

41. dan815

and terms are going up at n*(n+1) so

42. dan815

divided by 2

43. dan815

have to know the sum of squares formula basically

44. dan815

|dw:1434620893662:dw|

45. dan815

|dw:1434620949098:dw|

46. dan815

oops wrong wait!!

47. dan815

i think i have ot subtract sum of series from that

48. dan815

|dw:1434621049221:dw|

49. dan815

|dw:1434621081983:dw|

50. dan815

51. dan815

dont mind the scribbles before lol

52. mathmath333

wait lol

53. dan815

ok i go afk now

54. dan815

55. dan815

|dw:1434621259615:dw|

56. mathmath333

u have done it partially correct

57. dan815

maybe its subtract ugh xD

58. dan815

subtracting works i think

59. dan815

lemme just make it into a program i cant keep reyping that

60. mathmath333

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}}

61. dan815

:)

62. dan815

okay ill show u what i did its simple first i looked at this picture

63. dan815

|dw:1434621644190:dw|

64. dan815

we need to see the last term on the 20th line

65. dan815

and for every line we get 1 new term

66. dan815

so its the sum of increasing number by 1

67. dan815

um ya the main thing is to just find the nth term u are dealing with

68. dan815

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

69. dan815

that is the number we are adding +1 +2+3... to +19

70. dan815

i feel like im failing at explaining this lol, it should be simple xD

71. mathmath333

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

72. dan815

nice

73. dan815

thats a pretty different method from me

74. dan815

i did not notice that triangular number pattern there :O cool :)

75. mathmath333

i thought u did it that way

76. dan815

hehe no its a bit different

77. dan815

but its interting the triangular numbers came up anayway

78. dan815

interesting*