Question Number 3628 by madscientist last updated on 16/Dec/15
$${if}\:{the}\:{fibonacci}\:{sequence}\:{is}\: \\ $$$$\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8},\mathrm{13},\mathrm{21},\mathrm{34}… \\ $$$${where}\:{it}\:{is}\:\mathrm{1}+\mathrm{1}=\mathrm{2},\:\mathrm{2}+\mathrm{3}=\mathrm{5},\:\mathrm{3}+\mathrm{5}=\mathrm{8},… \\ $$$${how}\:{can}\:{we}\:{represent}\:{this}\:{sequence}\: \\ $$$${in}\:{summitation}\:{notation}\:\Sigma \\ $$$${or}\:{product}\:{notation}\:\Pi? \\ $$
Commented by 123456 last updated on 16/Dec/15
$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{there}\:\mathrm{also}\:\mathrm{a}\:\mathrm{explict}\:\mathrm{form}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of} \\ $$$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{and}\:\Phi=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 16/Dec/15
$${a}_{{n}} =\frac{\emptyset^{{n}} −\Phi^{{n}} }{\:\sqrt{\mathrm{5}}} \\ $$
Answered by Filup last updated on 17/Dec/15
$$\mathrm{Fibonacci}\:\mathrm{sequence}: \\ $$$${T}_{{n}} ={T}_{{n}−\mathrm{1}} +{T}_{{n}−\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{common}\:\mathrm{ratio}: \\ $$$${r}=\phi=\frac{{T}_{{n}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} } \\ $$$$\frac{{T}_{{n}−\mathrm{1}} +{T}_{{n}−\mathrm{2}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} } \\ $$$$\mathrm{1}+\frac{{T}_{{n}−\mathrm{2}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} } \\ $$$$\therefore\mathrm{1}+\frac{\mathrm{1}}{\phi}=\phi \\ $$$$\phi^{\mathrm{2}} −\phi−\mathrm{1}=\mathrm{0} \\ $$$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\Phi=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$${n}\mathrm{th}\:\mathrm{fibonacci}\:\mathrm{number}: \\ $$$$\phi_{{n}} =\frac{\phi^{{n}} −\Phi^{{n}} }{\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\therefore{Sum}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\phi_{{i}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\phi^{{i}} −\Phi^{{i}} }{\:\sqrt{\mathrm{5}}} \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
$$\mathrm{Find}\:\mathrm{common}\:\mathrm{ratio}: \\ $$$${r}=\phi=\frac{{T}_{{n}} }{{T}_{{n}−\mathrm{1}} }=\frac{{T}_{{n}−\mathrm{1}} }{{T}_{{n}−\mathrm{2}} }\:\:? \\ $$$$\frac{{T}_{\mathrm{3}} }{{T}_{\mathrm{2}} }=\:\frac{{T}_{\mathrm{2}} }{{T}_{\mathrm{1}} }? \\ $$$${T}_{\mathrm{1}} ={T}_{\mathrm{2}} =\mathrm{1}\:,\:{T}_{\mathrm{3}} =\mathrm{2} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{1}}? \\ $$
Commented by Filup last updated on 17/Dec/15
$$''{ratio}''\:\mathrm{isn}'\mathrm{t}\:\mathrm{a}\:\mathrm{good}\:\mathrm{word}\:\mathrm{to}\:\mathrm{use}.\:\mathrm{Simply}, \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{approaches} \\ $$$$\mathrm{as}\:{n}\rightarrow\infty \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{belive}\:\mathrm{that}\:{T}_{{n}} \phi\approx{T}_{{n}+\mathrm{1}} \:\mathrm{for}\:\mathrm{large}\:{n} \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
$$\mathcal{TH}\alpha\mathcal{NK}^{\mathcal{S}} ! \\ $$