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Question Number 81446 by john santu last updated on 13/Feb/20
given a_1 = 2 , a_2 = 3 and a_(n+2) = a_(n+1) + (a_n /2) find a_n =?
$$\mathrm{given}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{2}\:,\:\mathrm{a}_{\mathrm{2}} \:=\:\mathrm{3}\:\mathrm{and} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:+\:\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \:=? \\ $$
Commented by mr W last updated on 13/Feb/20
q^2 −q−(1/2)=0 q=((1±(√3))/2) a_n =A(((1+(√3))/2))^n +B(((1−(√3))/2))^n a_1 =A(((1+(√3))/2))+B(((1−(√3))/2))=2 a_2 =A(((1+(√3))/2))^2 +B(((1−(√3))/2))^2 =3 ⇒A+B=2 ⇒A−B=((2(√3))/3) ⇒A=((3+(√3))/3) ⇒B=((3−(√3))/3) ⇒a_n =(((3+(√3))/3))(((1+(√3))/2))^n +(((3−(√3))/3))(((1−(√3))/2))^n
$${q}^{\mathrm{2}} −{q}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${q}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}_{\mathrm{1}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$${a}_{\mathrm{2}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow{A}+{B}=\mathrm{2} \\ $$$$\Rightarrow{A}−{B}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow{a}_{{n}} =\left(\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$
Commented by behi83417@gmail.com last updated on 13/Feb/20
a_n =(1/( (√3)))[(((1+(√3))/2))^(n+1) − (((1−(√3))/2))^(n+1) ]
$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} −\:\:\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \:\right] \\ $$
Commented by jagoll last updated on 13/Feb/20
from a_n = (((3+(√3))/3))(((1+(√3))/2))^n +(((3−(√3))/3))(((1−(√3))/2))^n a_1 =(((3+(√3))/3))(((1+(√3))/2))+(((3−(√3))/3))(((1−(√3))/2)) = ((6+4(√3)+6−4(√3))/6) = ((12)/6) = 2 amazing sir
$$\mathrm{from}\:\mathrm{a}_{\mathrm{n}} \:=\:\left(\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{1}} \:=\left(\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\left(\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\:\frac{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{6}−\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{6}}\:=\:\frac{\mathrm{12}}{\mathrm{6}}\:=\:\mathrm{2} \\ $$$$\mathrm{amazing}\:\mathrm{sir} \\ $$
Commented by john santu last updated on 13/Feb/20
in generally if ⇒2a_(n+2) = 3a_(n+1) −(a_n /3) ⇒ 2q^2 −3q+(1/3)=0 6q^2 −9q+1 = 0 q = ((9 ± (√(81−24)))/(12)) = ((9 ± (√(57)))/(12)) a_(n ) = A(((9−(√(57)))/(12)))^n + B(((9+(√(57)))/(12)))^(n )
$$\mathrm{in}\:\mathrm{generally}\: \\ $$$$\mathrm{if}\:\Rightarrow\mathrm{2a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{3a}_{\mathrm{n}+\mathrm{1}} \:−\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{2q}^{\mathrm{2}} \:−\mathrm{3q}+\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{6q}^{\mathrm{2}} \:−\mathrm{9q}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{q}\:=\:\frac{\mathrm{9}\:\pm\:\sqrt{\mathrm{81}−\mathrm{24}}}{\mathrm{12}}\:=\:\frac{\mathrm{9}\:\pm\:\sqrt{\mathrm{57}}}{\mathrm{12}} \\ $$$$\mathrm{a}_{\mathrm{n}\:} =\:\mathrm{A}\left(\frac{\mathrm{9}−\sqrt{\mathrm{57}}}{\mathrm{12}}\right)^{\mathrm{n}} +\:\mathrm{B}\left(\frac{\mathrm{9}+\sqrt{\mathrm{57}}}{\mathrm{12}}\right)^{\mathrm{n}\:} \\ $$
Commented by jagoll last updated on 13/Feb/20
thank you mister W. i have the same problem with this kind.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister}\:\mathrm{W}.\:\mathrm{i}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{problem}\:\mathrm{with}\:\mathrm{this}\:\mathrm{kind}. \\ $$$$ \\ $$

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