Question

Identify a reasonable pattern and find a formula for \(n\) th term of the sequence :

\[\large -3,~21,~3,~129,~147,\ldots\]

Answer 1

we can build a degree 4 (maybe?) polynomial :)

Answer 2

haha sure guess we can do that with any sequence, including
\[1,~1,~2,~3,~5,~8,~13,~\cdots\]

;p

Answer 3

i actually reworked the closed form of the fibber today :)

Answer 4

the sequence in question is fibonacci like one..
but identifying the pattern might be tough as there is no uniq answer yeah

Answer 5

if i toss in a switch function, then for all points in the ... it can be zero or even some other function to turn on ;)

Answer 6

so your pattern is what, an alternating sign fibber?

Answer 7

well, not precise

Answer 8

was hoping to work it using
\[u_n = Au_{n-1}+Bu_{n-2}\]

where \(A\) and \(B\) are constants

Answer 9

what have you determined to be your recursive form?

Answer 10

exercise 15B / last problem
http://www.cimt.plymouth.ac.uk/projects/mepres/alevel/discrete_ch15.pdf

Answer 11

\[u_n = u_{n-1}+6u_{n-2}\]

http://www.wolframalpha.com/input/?i=solve+3+%3D+21x-3y%2C+129%3D3x%2B21y

Answer 12

ok, you already have the recursion ... i just figured it out lol

Answer 13

-3 21 3 129 147

-3x + 21y = 3
21x + 3y = 129

6x+y

6(3)+129 = 147

Answer 14

might be easier to work like this tho

f(n+2) = 6 f(n+1) + f(n)

f(n+2) - 6 f(n+1) - f(n) = 0

let f(n) = x^n

x^(n+2) - 6 x^(n+1) - x^(n) = 0

x^(n) (x^2 - 6x - 1) = 0

what are the roots of our quadratic?

Answer 15

i think my 6 is in the wrong spot tho ... didnt double check to see if my fingers were behaving

Answer 16

f(n+2) = f(n+1) + 6f(n)

right

Answer 17

x^2 - x - 6 = 0

x = 3, -2

Answer 18

since 3^n and (-2)^n are linearly independant, then we can form a ... linear combination of the solutions as:

A3^n + B(-2)^n and setup a system of 2 equations in 2 unknowns

do we like to start with n=0 or n=1?

Answer 19

0 i guess, the trick part is identifying a pattern and after that finding the general solution is pretty straightforward!

Answer 20

:) yeah

A+B = -3
3A -2B = 21

Answer 21

2A+2B = -6
3A -2B = 21
-----------
5A = 15

A = 3, so B = -9

Answer 22

we can chk with the wolf:

table [3(3^n)-9(-2)^n,{n,10}]

Answer 23

*B = -6
\[u_n = 3^{n+1}-6(-2)^{n}\]

Answer 24

lol ... fine we will use -6 instead :)

Answer 25

... wow, he can solve a closed form for some random fibber sequence, but that addition really throws him for a loop ;)

Answer 26

checks out wid wolf !
http://www.wolframalpha.com/input/?i=Table%5B3%5E%28n%2B1%29-6%28-2%29%5En%2C+%7Bn%2C0%2C4%7D%5D

i use to do these problems using generating functions, they're long but they gave nice practice with power series

Answer 27

i cant say that i have played with generating functions much, if at all.

Answer 28

is it unique ?

Answer 29

well i can tell you using -9 doesnt work .. so im gonna say yes :)

Answer 30

well :P

Answer 31

using generating functions

\[\begin{align}G(x) &= u_0+u_1x+u_2x^2+u_3x^3+\cdots\\~\\
&=-3 + 21x + (u_1+6u_0)x^2 + (u_2+6u_1)x^3 + \cdots \\~\\
&=-3+21x+x(u_1x+u_2x^2+\cdots) + 6x^2(u_0+u_1x+\cdots)\\~\\
&=-3+21x+x(G(x)-u_0) + 6x^2G(x)\\~\\
\end{align}\]
\(\implies G(x)[1-x-6x^2]=-3+21x-u_0x\\~\\
\begin{align}\implies G(x) &= \dfrac{-3+24x}{1-x-6x^2} \\~\\
&= \dfrac{3}{1-3x}-\dfrac{6}{1+2x}\\~\\
&=3\sum\limits_{n=0}^{\infty} (3x)^n-6\sum\limits_{n=0}^{\infty} (-2x)^n\\~\\
&=\sum\limits_{n=0}^{\infty} \left[\color{blue}{3^{n+1}-6(-2)^n}\right] x^n

\end{align}\)