if-the-fibonacci-sequence-is-1-1-2-3-5-8-13-21-34-where-it-is-1-1-2-2-3-5-3-5-8-how-can-we-represent-this-sequence-in-summitation-notation-or-product-notation-

Question Number 3628 by madscientist last updated on 16/Dec/15

i f t h e f i b o n a c c i s e q u e n c e i s

1, 1, 2, 3, 5, 8, 13, 21, 34 \dots

w h e r e i t i s 1 + 1 = 2, 2 + 3 = 5, 3 + 5 = 8, \dots

h o w c a n w e r e p r e s e n t t h i s s e q u e n c e

i n s u m m i t a t i o n n o t a t i o n Σ

o r p r o d u c t n o t a t i o n Π ?

Commented by 123456 last updated on 16/Dec/15

a_{n} = a_{n - 1} + a_{n - 2}

a_{1} = 1

a_{2} = 1

there also a explict form in terms of

ϕ = \frac{1 + \sqrt{5}}{2} and Φ = \frac{1 - \sqrt{5}}{2}

Commented by prakash jain last updated on 16/Dec/15

a_{n} = \frac{\emptyset^{n} - Φ^{n}}{\sqrt{5}}

Answered by Filup last updated on 17/Dec/15

Fibonacci sequence :

T_{n} = T_{n - 1} + T_{n - 2}

Find common ratio :

r = ϕ = \frac{T_{n}}{T_{n - 1}} = \frac{T_{n - 1}}{T_{n - 2}}

\frac{T_{n - 1} + T_{n - 2}}{T_{n - 1}} = \frac{T_{n - 1}}{T_{n - 2}}

1 + \frac{T_{n - 2}}{T_{n - 1}} = \frac{T_{n - 1}}{T_{n - 2}}

∴ 1 + \frac{1}{ϕ} = ϕ

ϕ^{2} - ϕ - 1 = 0

ϕ = \frac{1 + \sqrt{5}}{2} Φ = \frac{1 - \sqrt{5}}{2}

n th fibonacci number :

ϕ_{n} = \frac{ϕ^{n} - Φ^{n}}{\sqrt{5}}

∴ S u m = \underset{i = 1}{\sum^{n}} ϕ_{i} = \underset{i = 1}{\sum^{n}} \frac{ϕ^{i} - Φ^{i}}{\sqrt{5}}

Commented by Rasheed Soomro last updated on 17/Dec/15

Find common ratio :

r = ϕ = \frac{T_{n}}{T_{n - 1}} = \frac{T_{n - 1}}{T_{n - 2}} ?

\frac{T_{3}}{T_{2}} = \frac{T_{2}}{T_{1}} ?

T_{1} = T_{2} = 1, T_{3} = 2

\frac{2}{1} = \frac{1}{1} ?

Commented by Filup last updated on 17/Dec/15

^{″} {r a t i o}^{″} {isn}^{'} t a good word to use . Simply,

{it}^{'} s the ratio that the sequence approaches

as n \to \infty

I belive that T_{n} ϕ \approx T_{n + 1} for large n

Commented by Rasheed Soomro last updated on 17/Dec/15

TH α {NK}^{S}!