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Question Number 3628 by madscientist last updated on 16/Dec/15
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 Σ or product notation Π?
$${if}\:{the}\:{fibonacci}\:{sequence}\:{is}\: \\ $$$$\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8},\mathrm{13},\mathrm{21},\mathrm{34}… \\ $$$${where}\:{it}\:{is}\:\mathrm{1}+\mathrm{1}=\mathrm{2},\:\mathrm{2}+\mathrm{3}=\mathrm{5},\:\mathrm{3}+\mathrm{5}=\mathrm{8},… \\ $$$${how}\:{can}\:{we}\:{represent}\:{this}\:{sequence}\: \\ $$$${in}\:{summitation}\:{notation}\:\Sigma \\ $$$${or}\:{product}\:{notation}\:\Pi? \\ $$
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=((1+(√5))/2) and Φ=((1−(√5))/2)
$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{there}\:\mathrm{also}\:\mathrm{a}\:\mathrm{explict}\:\mathrm{form}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of} \\ $$$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{and}\:\Phi=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 16/Dec/15
a_n =((∅^n −Φ^n )/( (√5)))
$${a}_{{n}} =\frac{\emptyset^{{n}} −\Phi^{{n}} }{\:\sqrt{\mathrm{5}}} \\ $$
Answered by Filup last updated on 17/Dec/15
Fibonacci sequence: T_n =T_(n−1) +T_(n−2) Find common ratio: r=φ=(T_n /T_(n−1) )=(T_(n−1) /T_(n−2) ) ((T_(n−1) +T_(n−2) )/T_(n−1) )=(T_(n−1) /T_(n−2) ) 1+(T_(n−2) /T_(n−1) )=(T_(n−1) /T_(n−2) ) ∴1+(1/φ)=φ φ^2 −φ−1=0 φ=((1+(√5))/2) Φ=((1−(√5))/2) nth fibonacci number: φ_n =((φ^n −Φ^n )/( (√5))) ∴Sum=Σ_(i=1) ^n φ_i =Σ_(i=1) ^n ((φ^i −Φ^i )/( (√5)))
$$\mathrm{Fibonacci}\:\mathrm{sequence}: \\ $$$${T}_{{n}} ={T}_{{n}−\mathrm{1}} +{T}_{{n}−\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{common}\:\mathrm{ratio}: \\ $$$${r}=\phi=\frac{{T}_{{n}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} } \\ $$$$\frac{{T}_{{n}−\mathrm{1}} +{T}_{{n}−\mathrm{2}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} } \\ $$$$\mathrm{1}+\frac{{T}_{{n}−\mathrm{2}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} } \\ $$$$\therefore\mathrm{1}+\frac{\mathrm{1}}{\phi}=\phi \\ $$$$\phi^{\mathrm{2}} −\phi−\mathrm{1}=\mathrm{0} \\ $$$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\Phi=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$${n}\mathrm{th}\:\mathrm{fibonacci}\:\mathrm{number}: \\ $$$$\phi_{{n}} =\frac{\phi^{{n}} −\Phi^{{n}} }{\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\therefore{Sum}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\phi_{{i}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\phi^{{i}} −\Phi^{{i}} }{\:\sqrt{\mathrm{5}}} \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
Find common ratio: r=φ=(T_n /T_(n−1) )=(T_(n−1) /T_(n−2) ) ? (T_3 /T_2 )= (T_2 /T_1 )? T_1 =T_2 =1 , T_3 =2 (2/1)=(1/1)?
$$\mathrm{Find}\:\mathrm{common}\:\mathrm{ratio}: \\ $$$${r}=\phi=\frac{{T}_{{n}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} }\:\:? \\ $$$$\frac{{T}_{\mathrm{3}} }{{T}_{\mathrm{2}} }=\:\frac{{T}_{\mathrm{2}} }{{T}_{\mathrm{1}} }? \\ $$$${T}_{\mathrm{1}} ={T}_{\mathrm{2}} =\mathrm{1}\:,\:{T}_{\mathrm{3}} =\mathrm{2} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{1}}? \\ $$
Commented by Filup last updated on 17/Dec/15
′′ratio′′ isn′t a good word to use. Simply, it′s the ratio that the sequence approaches as n→∞ I belive that T_n φ≈T_(n+1) for large n
$$''{ratio}''\:\mathrm{isn}'\mathrm{t}\:\mathrm{a}\:\mathrm{good}\:\mathrm{word}\:\mathrm{to}\:\mathrm{use}.\:\mathrm{Simply}, \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{approaches} \\ $$$$\mathrm{as}\:{n}\rightarrow\infty \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{belive}\:\mathrm{that}\:{T}_{{n}} \phi\approx{T}_{{n}+\mathrm{1}} \:\mathrm{for}\:\mathrm{large}\:{n} \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
THαNK^S !
$$\mathcal{TH}\alpha\mathcal{NK}^{\mathcal{S}} ! \\ $$

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