mathmath333
  • mathmath333
Fun question
Mathematics
katieb
  • katieb
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}& \normalsize \text{Find the 20th term of the sequence }\hspace{.33em}\\~\\ &1,\ 5,\ 15,\ 34,\ 65\cdots\cdots \end{align}}\)
ikram002p
  • ikram002p
the magic lol
ikram002p
  • ikram002p
do u really know how does the magic squares works ? it would be fun to play with

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mathmath333
  • mathmath333
idk magic squares
ikram002p
  • ikram002p
oh so u just know the formula that defines the suitable sum xD nvm
mathmath333
  • mathmath333
no i dont know any formula for this nth term
ikram002p
  • ikram002p
:) lets see if someone else know
anonymous
  • anonymous
it s 4010 and here is the formula a(n) = n*(n^2 + 1)/2.
mathmath333
  • mathmath333
there may be the predefined nth term , but thats not any fun doing this
anonymous
  • anonymous
@mathmath333 what do you think ?
mathmath333
  • mathmath333
i think we should do any observation
anonymous
  • anonymous
so enjoy observing :)
anonymous
  • 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
mathmath333
  • mathmath333
how did u know the formula by the way
ikram002p
  • 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 :)
mathmath333
  • mathmath333
that looks little cumbersome
anonymous
  • 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
ikram002p
  • ikram002p
well this is one way to do it -.- u can make sine cos combination (fourier)
ikram002p
  • ikram002p
anyway this is wolfram answer for the new polynomial (as u dont wanna the predefined )
anonymous
  • anonymous
@ikram002p gave you the normal right method to follow when dealing with those kind of sequence
ikram002p
  • ikram002p
http://www.wolframalpha.com/input/?i=a%2Bb%2Bc%2Bd%2B1%3D5%2C16a%2B8b%2B4c%2B2d%2B1%3D15%2C81a%2B27b%2B9c%2B3d%2B1%3D34%2C256a%2B64b%2B16c%2B4d%2B1%3D65+
mathmath333
  • mathmath333
ok a hint \(\large \color{black}{\begin{align} &1,\\~\\&\ 2+3,\\~\\&\ 4+5+6,\\~\\&\ 7+8+9+10,\\~\\&\ 11+12+13+14+15 \\~\\&\cdots\cdots \\~\\ \end{align}}\)
ikram002p
  • ikram002p
this is what been defined :O
dan815
  • dan815
oh simple
ikram002p
  • ikram002p
-.-
dan815
  • dan815
they just seem so close to doubling lol
dan815
  • dan815
|dw:1434620107872:dw|
dan815
  • dan815
|dw:1434620166190:dw|
dan815
  • dan815
|dw:1434620264189:dw|
dan815
  • dan815
solve a,b,c and u will ahve an equation that can get any xth term T(x)=...
dan815
  • dan815
I dont personally like this method though, we havent probven its a diff of 3 from there :)
dan815
  • dan815
but then again the equation u get does saitsfy all the terms uare given, so u cannot say its wrong
dan815
  • dan815
but then again the argument is that, there is obviously going to be some polynomial equation taht will will any given finite sequence,
dan815
  • 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
mathmath333
  • mathmath333
ok i will give a last hint
dan815
  • dan815
WAIT !! i solve it, u want me to do it another way!! dont tell me
mathmath333
  • mathmath333
ok
dan815
  • dan815
1 5 15 34 65 . . . 1 2+3 4+5+6 7+8+9+10
dan815
  • dan815
-.-
dan815
  • dan815
oaky so all we gotta do is calculate the nth term we will see at 20
dan815
  • dan815
and terms are going up at n*(n+1) so
dan815
  • dan815
divided by 2
dan815
  • dan815
have to know the sum of squares formula basically
dan815
  • dan815
|dw:1434620893662:dw|
dan815
  • dan815
|dw:1434620949098:dw|
dan815
  • dan815
oops wrong wait!!
dan815
  • dan815
i think i have ot subtract sum of series from that
dan815
  • dan815
|dw:1434621049221:dw|
dan815
  • dan815
|dw:1434621081983:dw|
dan815
  • dan815
is that the answer??
dan815
  • dan815
dont mind the scribbles before lol
mathmath333
  • mathmath333
wait lol
dan815
  • dan815
ok i go afk now
dan815
  • dan815
wait its add! omg
dan815
  • dan815
|dw:1434621259615:dw|
mathmath333
  • mathmath333
u have done it partially correct
dan815
  • dan815
maybe its subtract ugh xD
dan815
  • dan815
subtracting works i think
dan815
  • dan815
lemme just make it into a program i cant keep reyping that
mathmath333
  • mathmath333
well that works now this one , m still confuzed on how u got that but its correct \(\large \color{black}{\begin{align} k=\dfrac{n^2(n+1)}{2}-\dfrac{n(n-1)}{2 }\hspace{.33em}\\~\\ \end{align}}\)
dan815
  • dan815
:)
dan815
  • dan815
okay ill show u what i did its simple first i looked at this picture
dan815
  • dan815
|dw:1434621644190:dw|
dan815
  • dan815
we need to see the last term on the 20th line
dan815
  • dan815
and for every line we get 1 new term
dan815
  • dan815
so its the sum of increasing number by 1
dan815
  • dan815
um ya the main thing is to just find the nth term u are dealing with
dan815
  • 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
dan815
  • dan815
that is the number we are adding +1 +2+3... to +19
dan815
  • dan815
i feel like im failing at explaining this lol, it should be simple xD
mathmath333
  • 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
dan815
  • dan815
nice
dan815
  • dan815
thats a pretty different method from me
dan815
  • dan815
i did not notice that triangular number pattern there :O cool :)
mathmath333
  • mathmath333
i thought u did it that way
dan815
  • dan815
hehe no its a bit different
dan815
  • dan815
but its interting the triangular numbers came up anayway
dan815
  • dan815
interesting*

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