Quadratic Nth Term Sequences?

Question

Hey.

I would class my self as very good at the linear sequences but not so much with regard to the quadratic sequences. I was wondering if someone could please help me.

Can you explain how to solve quadratic nth term sequences? Their also called the changing difference type i believe. For the linear sequences i use the formula dn+ (a-d). Also what is the fibonacci sequence and what does it have to do with the Nth term?

All comments are very much appreciated. Many thanks!

Answer 1

I knows little about sequence, I don't quite understand what linear

and quadratic sequence means. I had learnt the Fibonacci

sequences before and know that this sequence has a special

reccrrence relation governing its terms. Somebody believe that

it conceives some "mystery" of the Universe. This relation is

xn=xn-1+xn-2 ,where n>2 and xn is the nth term. And the sequ.

is 1,1,2,3,5,8,13,21,......xi...

let xn=k^n where k is a real number,then

xn=xn-1+xn-2=>

k^n=k^(n-1)+k^(n-2)=>

k^2=k+1=>

k^2-k-1=0 (is it the "quadratic" you mean?)

solve for k,get

k=(1+5^1/2)/2 or k=(1-5^1/2)/2,

so the gen.solution for xi is

xi=A[(1+5^1/2)/2]^i+B[(1-5^1/2)/2]^i, where A,B are constants

Take x1=x2=1,A,B can be found to be A=1/5^1/2 and B=-1/5^1/2.

The strange thing apparently is that xi which contains irrational nos.

turns out to be an +ve integer.

Answer 2

well i can just tell u what is fibonacci series as far as i remember

the series is

0, 1 ,1 ,2 ,3 ,5 ,8 ,13 ,21 ,34

here the series starts with 0 and 1

then the third term is the sum of first and second term

then the fourth term is sum of second and third term

then the fifth term is sum of third and fourth term so on and so forth.

sorry but i can't help u with the rest part of your question