Menu Close

The-first-term-of-a-sequence-is-1-the-second-is-2-and-every-term-is-the-sum-of-the-two-preceding-terms-The-n-th-term-is-

Tinku Tara 0 Comments
Pin




Question Number 25381 by Tinkutara last updated on 09/Dec/17
The first term of a sequence is 1, the second is 2 and every term is the sum of the two preceding terms. The n^(th) term is.
$$\mathrm{The}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{1},\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{is}\:\mathrm{2}\:\mathrm{and}\:\mathrm{every}\:\mathrm{term}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{preceding}\:\mathrm{terms}.\:\mathrm{The}\:{n}^{\mathrm{th}} \:\mathrm{term} \\ $$$$\mathrm{is}. \\ $$
Answered by prakash jain last updated on 09/Dec/17
a_n =a_(n−1) +a_(n−2) characteristic equation x^2 −x−1=0 λ_1 =((1+(√5))/2), λ_2 =((1−(√5))/2) a_n =c_1 (((1+(√5))/2))^n +c_2 (((1−(√5))/2))^n n=0,a_0 =1 c_1 +c_2 =1 (i) n=1,a_1 =2 c_1 ((1+(√5))/2)+c_2 ((1−(√5))/2)=2 (√5)c_1 =2−((1−(√5))/2)⇒c_1 =((3+(√5))/(2(√5))) c_2 =1−c_1 =(((√5)−3)/(2(√5))) a_n =(((3+(√5))/(2(√5))))(((1+(√5))/2))^n +((((√5)−3)/(2(√5))))(((1−(√5))/2))^n
$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \\ $$$${characteristic}\:{equation} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\lambda_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\:\lambda_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={c}_{\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{c}_{\mathrm{2}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${n}=\mathrm{0},{a}_{\mathrm{0}} =\mathrm{1} \\ $$$${c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\mathrm{1}\:\:\:\left({i}\right) \\ $$$${n}=\mathrm{1},{a}_{\mathrm{1}} =\mathrm{2} \\ $$$${c}_{\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+{c}_{\mathrm{2}} \frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{2} \\ $$$$\sqrt{\mathrm{5}}{c}_{\mathrm{1}} =\mathrm{2}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{c}_{\mathrm{1}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$${c}_{\mathrm{2}} =\mathrm{1}−{c}_{\mathrm{1}} =\frac{\sqrt{\mathrm{5}}−\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$${a}_{{n}} =\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\sqrt{\mathrm{5}}−\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{5}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$
Commented by Tinkutara last updated on 09/Dec/17
What is characteristic equation? And why a_n =c_1 λ_1 ^n +c_2 λ_2 ^n ?
$${What}\:{is}\:{characteristic}\:{equation}?\:{And} \\ $$$${why}\:{a}_{{n}} ={c}_{\mathrm{1}} \lambda_{\mathrm{1}} ^{{n}} +{c}_{\mathrm{2}} \lambda_{\mathrm{2}} ^{{n}} ? \\ $$
Commented by prakash jain last updated on 09/Dec/17
x^2 −x−1=0 λ_1 ^2 −λ_1 −1=0 λ_2 ^2 −λ_2 −1=0 c_1 λ_1 ^(n−2) (λ_1 ^2 −λ_1 −1)+c_2 λ_2 ^(n−2) (λ_2 ^2 −λ_2 −1)=0 ⇒(c_1 λ_1 ^n +c_2 λ_2 ^n )=(c_1 λ_1 ^(n−1) +c_2 λ_2 ^(n−1) )+(c_1 λ_1 ^(n−2) +c_2 λ_2 ^(n−2) ) ⇒a_n =a_(n−1) +a_(n−2) The general solution of the equation is a_n =c_1 λ_1 ^n +c_2 λ_2 ^n This is the general solution when λ_1 ≠λ_2 when λ_1 =λ_2 =λ a_n =(c_1 +c_2 n)λ^n To know more read linear difference equation
$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\lambda_{\mathrm{1}} ^{\mathrm{2}} −\lambda_{\mathrm{1}} −\mathrm{1}=\mathrm{0} \\ $$$$\lambda_{\mathrm{2}} ^{\mathrm{2}} −\lambda_{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${c}_{\mathrm{1}} \lambda_{\mathrm{1}} ^{{n}−\mathrm{2}} \left(\lambda_{\mathrm{1}} ^{\mathrm{2}} −\lambda_{\mathrm{1}} −\mathrm{1}\right)+{c}_{\mathrm{2}} \lambda_{\mathrm{2}} ^{{n}−\mathrm{2}} \left(\lambda_{\mathrm{2}} ^{\mathrm{2}} −\lambda_{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({c}_{\mathrm{1}} \lambda_{\mathrm{1}} ^{{n}} +{c}_{\mathrm{2}} \lambda_{\mathrm{2}} ^{{n}} \right)=\left({c}_{\mathrm{1}} \lambda_{\mathrm{1}} ^{{n}−\mathrm{1}} +{c}_{\mathrm{2}} \lambda_{\mathrm{2}} ^{{n}−\mathrm{1}} \right)+\left({c}_{\mathrm{1}} \lambda_{\mathrm{1}} ^{{n}−\mathrm{2}} +{c}_{\mathrm{2}} \lambda_{\mathrm{2}} ^{{n}−\mathrm{2}} \right) \\ $$$$\Rightarrow{a}_{{n}} ={a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{is}\:{a}_{{n}} ={c}_{\mathrm{1}} \lambda_{\mathrm{1}} ^{{n}} +{c}_{\mathrm{2}} \lambda_{\mathrm{2}} ^{{n}} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{when} \\ $$$$\lambda_{\mathrm{1}} \neq\lambda_{\mathrm{2}} \\ $$$$\mathrm{when}\:\lambda_{\mathrm{1}} =\lambda_{\mathrm{2}} =\lambda \\ $$$${a}_{{n}} =\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {n}\right)\lambda^{{n}} \\ $$$$\mathrm{To}\:\mathrm{know}\:\mathrm{more} \\ $$$$\mathrm{read}\:\mathrm{linear}\:\mathrm{difference}\:\mathrm{equation} \\ $$
Commented by prakash jain last updated on 09/Dec/17
Characteristic equation is what you get after replacing a_n by x^n .
$$\mathrm{Characteristic}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{what} \\ $$$$\mathrm{you}\:\mathrm{get}\:\mathrm{after}\:\mathrm{replacing}\:{a}_{{n}} \:\mathrm{by}\:{x}^{{n}} . \\ $$
Commented by Tinkutara last updated on 10/Dec/17
Can we apply these methods anywhere? For example, in AGP?
$${Can}\:{we}\:{apply}\:{these}\:{methods}\:{anywhere}? \\ $$$${For}\:{example},\:{in}\:{AGP}? \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *