A community for students.
Here's the question you clicked on:
 0 viewing
ganeshie8
 one year ago
Identify a reasonable pattern and find a formula for \(n\) th term of the sequence :
\[\large 3,~21,~3,~129,~147,\ldots\]
ganeshie8
 one year ago
Identify a reasonable pattern and find a formula for \(n\) th term of the sequence : \[\large 3,~21,~3,~129,~147,\ldots\]

This Question is Closed

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2we can build a degree 4 (maybe?) polynomial :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3haha sure guess we can do that with any sequence, including \[1,~1,~2,~3,~5,~8,~13,~\cdots\] ;p

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i actually reworked the closed form of the fibber today :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3the sequence in question is fibonacci like one.. but identifying the pattern might be tough as there is no uniq answer yeah

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2if i toss in a switch function, then for all points in the ... it can be zero or even some other function to turn on ;)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so your pattern is what, an alternating sign fibber?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3was hoping to work it using \[u_n = Au_{n1}+Bu_{n2}\] where \(A\) and \(B\) are constants

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2what have you determined to be your recursive form?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3exercise 15B / last problem http://www.cimt.plymouth.ac.uk/projects/mepres/alevel/discrete_ch15.pdf

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[u_n = u_{n1}+6u_{n2}\] http://www.wolframalpha.com/input/?i=solve+3+%3D+21x3y%2C+129%3D3x%2B21y

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2ok, you already have the recursion ... i just figured it out lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.23 21 3 129 147 3x + 21y = 3 21x + 3y = 129 6x+y 6(3)+129 = 147

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2might 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?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i think my 6 is in the wrong spot tho ... didnt double check to see if my fingers were behaving

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3f(n+2) = f(n+1) + 6f(n) right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3x^2  x  6 = 0 x = 3, 2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2since 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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.30 i guess, the trick part is identifying a pattern and after that finding the general solution is pretty straightforward!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2:) yeah A+B = 3 3A 2B = 21

amistre64
 one year ago
Best ResponseYou've already chosen the best response.22A+2B = 6 3A 2B = 21  5A = 15 A = 3, so B = 9

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2we can chk with the wolf: table [3(3^n)9(2)^n,{n,10}]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3*B = 6 \[u_n = 3^{n+1}6(2)^{n}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2lol ... fine we will use 6 instead :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2... wow, he can solve a closed form for some random fibber sequence, but that addition really throws him for a loop ;)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3checks out wid wolf ! http://www.wolframalpha.com/input/?i=Table%5B3%5E%28n%2B1%296%282%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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i cant say that i have played with generating functions much, if at all.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well i can tell you using 9 doesnt work .. so im gonna say yes :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3using generating functions \[\begin{align}G(x) &= u_0+u_1x+u_2x^2+u_3x^3+\cdots\\~\\ &=3 + 21x + (u_1+6u_0)x^2 + (u_2+6u_1)x^3 + \cdots \\~\\ &=3+21x+x(u_1x+u_2x^2+\cdots) + 6x^2(u_0+u_1x+\cdots)\\~\\ &=3+21x+x(G(x)u_0) + 6x^2G(x)\\~\\ \end{align}\] \(\implies G(x)[1x6x^2]=3+21xu_0x\\~\\ \begin{align}\implies G(x) &= \dfrac{3+24x}{1x6x^2} \\~\\ &= \dfrac{3}{13x}\dfrac{6}{1+2x}\\~\\ &=3\sum\limits_{n=0}^{\infty} (3x)^n6\sum\limits_{n=0}^{\infty} (2x)^n\\~\\ &=\sum\limits_{n=0}^{\infty} \left[\color{blue}{3^{n+1}6(2)^n}\right] x^n \end{align}\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.