Question

Fun question

Answer 1

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

Answer 2

the magic lol

Answer 3

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

Answer 4

idk magic squares

Answer 5

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

Answer 6

no i dont know any formula for this nth term

Answer 7

:) lets see if someone else know

Answer 8

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

Answer 9

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

Answer 10

@mathmath333 what do you think ?

Answer 11

i think we should do any observation

Answer 12

so enjoy observing :)

Answer 13

the first 20 term : 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010

Answer 14

how did u know the formula by the way

Answer 15

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 :)

Answer 16

that looks little cumbersome

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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+

Answer 22

ok a hint


\(\large \color{black}{\begin{align}
&1,\\~\\&\ 2+3,\\~\\&\ 4+5+6,\\~\\&\ 7+8+9+10,\\~\\&\ 11+12+13+14+15 \\~\\&\cdots\cdots \\~\\
\end{align}}\)

Answer 23

this is what been defined :O

Answer 24

oh simple

Answer 25

-.-

Answer 26

they just seem so close to doubling lol