Fun question

- mathmath333

Fun question

- katieb

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- 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

the magic lol

- ikram002p

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

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## More answers

- mathmath333

idk magic squares

- ikram002p

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

- mathmath333

no i dont know any formula for this nth term

- ikram002p

:) lets see if someone else know

- anonymous

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

- mathmath333

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

- anonymous

@mathmath333 what do you think ?

- mathmath333

i think we should do any observation

- anonymous

so enjoy observing :)

- 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

how did u know the formula by the way

- 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

that looks little cumbersome

- 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

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

- ikram002p

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

- anonymous

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

- 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

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

this is what been defined :O

- dan815

oh simple

- ikram002p

-.-

- dan815

they just seem so close to doubling lol

- dan815

|dw:1434620107872:dw|

- dan815

|dw:1434620166190:dw|

- dan815

|dw:1434620264189:dw|

- dan815

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

- dan815

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

- dan815

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

- 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

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

ok i will give a last hint

- dan815

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

- mathmath333

ok

- dan815

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

- dan815

-.-

- dan815

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

- dan815

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

- dan815

divided by 2

- dan815

have to know the sum of squares formula basically

- dan815

|dw:1434620893662:dw|

- dan815

|dw:1434620949098:dw|

- dan815

oops wrong wait!!

- dan815

i think i have ot subtract sum of series from that

- dan815

|dw:1434621049221:dw|

- dan815

|dw:1434621081983:dw|

- dan815

is that the answer??

- dan815

dont mind the scribbles before lol

- mathmath333

wait lol

- dan815

ok i go afk now

- dan815

wait its add! omg

- dan815

|dw:1434621259615:dw|

- mathmath333

u have done it partially correct

- dan815

maybe its subtract ugh xD

- dan815

subtracting works i think

- dan815

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

- 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

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

- dan815

|dw:1434621644190:dw|

- dan815

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

- dan815

and for every line we get 1 new term

- dan815

so its the sum of increasing number by 1

- dan815

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

- 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

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

- dan815

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

- 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

nice

- dan815

thats a pretty different method from me

- dan815

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

- mathmath333

i thought u did it that way

- dan815

hehe no its a bit different

- dan815

but its interting the triangular numbers came up anayway

- dan815

interesting*

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