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Question Number 87647 by john santu last updated on 05/Apr/20

If a_1 = 1 , a_(n+1 ) = 2a_n + 5 , n = 1,2,3,.... then a_(100) = ?

$$\mathrm{If}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:,\:\mathrm{a}_{\mathrm{n}+\mathrm{1}\:} =\:\mathrm{2a}_{\mathrm{n}} \:+\:\mathrm{5}\:,\:\mathrm{n}\:=\: \\ $$$$\mathrm{1},\mathrm{2},\mathrm{3},....\:\mathrm{then}\:\mathrm{a}_{\mathrm{100}} \:=\:? \\ $$

Answered by mr W last updated on 05/Apr/20

a_(n+1) =2a_n +5 let a_n =Ap^n +C a_(n+1) =Ap^(n+1) +C a_(n+1) =2a_n +5 Ap^(n+1) +C=2Ap^n +2C+5 ⇒C=2C+5 ⇒C=−5 ⇒Ap^(n+1) =2Ap^n ⇒p=2 ⇒a_n =A2^n −5 a_1 =2A−5=1 ⇒A=3 ⇒a_n =3×2^n −5 a_(100) =3×2^(100) −5

$${a}_{{n}+\mathrm{1}} =\mathrm{2}{a}_{{n}} +\mathrm{5} \\ $$$${let}\:{a}_{{n}} ={Ap}^{{n}} +{C} \\ $$$${a}_{{n}+\mathrm{1}} ={Ap}^{{n}+\mathrm{1}} +{C} \\ $$$${a}_{{n}+\mathrm{1}} =\mathrm{2}{a}_{{n}} +\mathrm{5} \\ $$$${Ap}^{{n}+\mathrm{1}} +{C}=\mathrm{2}{Ap}^{{n}} +\mathrm{2}{C}+\mathrm{5} \\ $$$$\Rightarrow{C}=\mathrm{2}{C}+\mathrm{5}\:\Rightarrow{C}=−\mathrm{5} \\ $$$$\Rightarrow{Ap}^{{n}+\mathrm{1}} =\mathrm{2}{Ap}^{{n}} \Rightarrow{p}=\mathrm{2} \\ $$$$\Rightarrow{a}_{{n}} ={A}\mathrm{2}^{{n}} −\mathrm{5} \\ $$$${a}_{\mathrm{1}} =\mathrm{2}{A}−\mathrm{5}=\mathrm{1}\:\Rightarrow{A}=\mathrm{3} \\ $$$$\Rightarrow{a}_{{n}} =\mathrm{3}×\mathrm{2}^{{n}} −\mathrm{5} \\ $$$${a}_{\mathrm{100}} =\mathrm{3}×\mathrm{2}^{\mathrm{100}} −\mathrm{5} \\ $$

Commented by john santu last updated on 05/Apr/20

why not valid the method x^2 −2x−5=0 x = ((2 ± 2(√6))/2) = 1 ±(√6) a_n = A(1+(√6))^n +B(1−(√6))^n

$$\mathrm{why}\:\mathrm{not}\:\mathrm{valid}\:\mathrm{the}\:\mathrm{method} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{2}\:\pm\:\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}}\:=\:\mathrm{1}\:\pm\sqrt{\mathrm{6}} \\ $$$$\mathrm{a}_{\mathrm{n}} \:=\:\mathrm{A}\left(\mathrm{1}+\sqrt{\mathrm{6}}\right)^{\mathrm{n}} +\mathrm{B}\left(\mathrm{1}−\sqrt{\mathrm{6}}\right)^{\mathrm{n}} \\ $$

Commented by mr W last updated on 05/Apr/20

because 5 is a constant, not the coef. of a xth term, in the recursive relation.

$${because}\:\mathrm{5}\:{is}\:{a}\:{constant},\:{not}\:{the}\:{coef}. \\ $$$${of}\:{a}\:{xth}\:{term},\:{in}\:{the}\:{recursive}\:{relation}. \\ $$

Commented by john santu last updated on 05/Apr/20

oo i understand sir.

$$\mathrm{oo}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}. \\ $$

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