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Question Number 56864 by Joel578 last updated on 25/Mar/19

Find the solution of recurrence relation a_n = 2a_(n−1) + 3a_(n−2) , with a_0 = 1, a_1 = 2

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{recurrence}\:\mathrm{relation} \\ $$$${a}_{{n}} \:=\:\mathrm{2}{a}_{{n}−\mathrm{1}} \:+\:\mathrm{3}{a}_{{n}−\mathrm{2}} ,\:\mathrm{with}\:{a}_{\mathrm{0}} \:=\:\mathrm{1},\:{a}_{\mathrm{1}} \:=\:\mathrm{2} \\ $$

Commented by maxmathsup by imad last updated on 25/Mar/19

⇒a_(n+2) =2a_(n+1) +3 a_n ⇒a_(n+2) −2a_(n+1) −3a_n =0 ⇒caracteristic equation is x^2 −2x−3 =0 ⇒Δ^′ =1+3=4 ⇒x_1 =1+2=3 and x_2 =1−2=−1 ⇒ a_n =α 3^n +β (−1)^n initial conditions a_0 =1=α+β and a_1 =2=3α −β ⇒α+β =1 and 3α−β=2 ⇒3α−(1−α)=2 ⇒ 4α =3 ⇒α=(3/4) ⇒β=1−(3/4) =(1/4) ⇒a_n =(3/4) 3^n +(((−1)^n )/4) ⇒ a_n =(1/4){3^(n+1) +(−1)^n }

$$\Rightarrow{a}_{{n}+\mathrm{2}} =\mathrm{2}{a}_{{n}+\mathrm{1}} \:+\mathrm{3}\:{a}_{{n}} \:\Rightarrow{a}_{{n}+\mathrm{2}} −\mathrm{2}{a}_{{n}+\mathrm{1}} −\mathrm{3}{a}_{{n}} =\mathrm{0}\:\Rightarrow{caracteristic}\:{equation}\:{is} \\ $$$${x}^{\mathrm{2}} \:−\mathrm{2}{x}−\mathrm{3}\:=\mathrm{0}\:\Rightarrow\Delta^{'} =\mathrm{1}+\mathrm{3}=\mathrm{4}\:\Rightarrow{x}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}=\mathrm{3}\:{and}\:{x}_{\mathrm{2}} =\mathrm{1}−\mathrm{2}=−\mathrm{1}\:\:\Rightarrow \\ $$$${a}_{{n}} =\alpha\:\mathrm{3}^{{n}} \:+\beta\:\left(−\mathrm{1}\right)^{{n}} \:\:\:{initial}\:{conditions} \\ $$$${a}_{\mathrm{0}} =\mathrm{1}=\alpha+\beta\:\:\:{and}\:\:{a}_{\mathrm{1}} =\mathrm{2}=\mathrm{3}\alpha\:−\beta\:\Rightarrow\alpha+\beta\:=\mathrm{1}\:{and}\:\mathrm{3}\alpha−\beta=\mathrm{2}\:\Rightarrow\mathrm{3}\alpha−\left(\mathrm{1}−\alpha\right)=\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{4}\alpha\:=\mathrm{3}\:\Rightarrow\alpha=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow\beta=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow{a}_{{n}} =\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{3}^{{n}} \:+\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}}\:\Rightarrow \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{3}^{{n}+\mathrm{1}} \:+\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$

Commented by Joel578 last updated on 26/Mar/19

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Commented by turbo msup by abdo last updated on 26/Mar/19

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by mr W last updated on 25/Mar/19

let a_n =Ap^n +Bq^n ⇒ Ap^n +Bq^n =2Ap^(n−1) +2Bq^(n−1) +3Ap^(n−2) +3Bq^(n−2) ⇒ A(p^2 −2p−3)p^(n−2) +B(q^2 −2q−3)q^(n−2) =0 ⇒p and q are roots of x^2 −2x−3=0 (x+1)(x−3)=0 ⇒p=−1 and q=3 ⇒a_n =A×(−1)^n +B×3^n ⇒a_0 =A+B=1 ...(i) ⇒a_1 =−A+3B=2 ...(ii) ⇒A=(1/4) ⇒B=(3/4) ⇒a_n =(1/4)[(−1)^n +3^(n+1) ]

$${let}\:{a}_{{n}} ={Ap}^{{n}} +{Bq}^{{n}} \\ $$$$\Rightarrow\:{Ap}^{{n}} +{Bq}^{{n}} =\mathrm{2}{Ap}^{{n}−\mathrm{1}} +\mathrm{2}{Bq}^{{n}−\mathrm{1}} +\mathrm{3}{Ap}^{{n}−\mathrm{2}} +\mathrm{3}{Bq}^{{n}−\mathrm{2}} \\ $$$$\Rightarrow\:{A}\left({p}^{\mathrm{2}} −\mathrm{2}{p}−\mathrm{3}\right){p}^{{n}−\mathrm{2}} +{B}\left({q}^{\mathrm{2}} −\mathrm{2}{q}−\mathrm{3}\right){q}^{{n}−\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}\:{and}\:{q}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=−\mathrm{1}\:{and}\:{q}=\mathrm{3} \\ $$$$\Rightarrow{a}_{{n}} ={A}×\left(−\mathrm{1}\right)^{{n}} +{B}×\mathrm{3}^{{n}} \\ $$$$\Rightarrow{a}_{\mathrm{0}} ={A}+{B}=\mathrm{1}\:\:\:...\left({i}\right) \\ $$$$\Rightarrow{a}_{\mathrm{1}} =−{A}+\mathrm{3}{B}=\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\left(−\mathrm{1}\right)^{{n}} +\mathrm{3}^{{n}+\mathrm{1}} \right] \\ $$

Commented by Joel578 last updated on 26/Mar/19

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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