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Given-a-n-1-3-6a-n-and-a-4-9-Find-a-n-

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Question Number 206997 by efronzo1 last updated on 03/May/24
Given a_(n+1) = 3−6a_n and a_4 =−9 Find a_n =?
$$\mathrm{Given}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\:\mathrm{3}−\mathrm{6a}_{\mathrm{n}} \:\mathrm{and}\:\mathrm{a}_{\mathrm{4}} =−\mathrm{9} \\ $$$$\:\mathrm{Find}\:\mathrm{a}_{\mathrm{n}} =?\: \\ $$
Answered by mr W last updated on 04/May/24
a_(n+1) +6a_n =3 let a_n =b_n +k b_(n+1) +k+6(b_n +k)=3 b_(n+1) +6b_n =3−7k=^! 0 ⇒k=(3/7) b_n =−6b_(n−1) =(−6)^2 b_(n−2) =...=(−6)^(n−4) b_4 a_n −k=(−6)^(n−4) (a_4 −k) a_n =(−6)^(n−4) (a_4 −k)+k a_n =(−6)^(n−4) (−9−(3/7))+(3/7) ⇒a_n =((11(−6)^(n−3) +3)/7)
$${a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} =\mathrm{3} \\ $$$${let}\:{a}_{{n}} ={b}_{{n}} +{k} \\ $$$${b}_{{n}+\mathrm{1}} +{k}+\mathrm{6}\left({b}_{{n}} +{k}\right)=\mathrm{3} \\ $$$${b}_{{n}+\mathrm{1}} +\mathrm{6}{b}_{{n}} =\mathrm{3}−\mathrm{7}{k}\overset{!} {=}\mathrm{0}\:\Rightarrow{k}=\frac{\mathrm{3}}{\mathrm{7}} \\ $$$${b}_{{n}} =−\mathrm{6}{b}_{{n}−\mathrm{1}} =\left(−\mathrm{6}\right)^{\mathrm{2}} {b}_{{n}−\mathrm{2}} =…=\left(−\mathrm{6}\right)^{{n}−\mathrm{4}} {b}_{\mathrm{4}} \\ $$$${a}_{{n}} −{k}=\left(−\mathrm{6}\right)^{{n}−\mathrm{4}} \left({a}_{\mathrm{4}} −{k}\right) \\ $$$${a}_{{n}} =\left(−\mathrm{6}\right)^{{n}−\mathrm{4}} \left({a}_{\mathrm{4}} −{k}\right)+{k} \\ $$$${a}_{{n}} =\left(−\mathrm{6}\right)^{{n}−\mathrm{4}} \left(−\mathrm{9}−\frac{\mathrm{3}}{\mathrm{7}}\right)+\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{11}\left(−\mathrm{6}\right)^{{n}−\mathrm{3}} +\mathrm{3}}{\mathrm{7}} \\ $$
Commented by efronzo1 last updated on 03/May/24
⇒a_(n+1) = b_(n+1) +k ⇒3−a_n = b_(n+1) +k ⇒3−(b_n +k)= b_(n+1) +k ⇒b_(n+1) +b_n = 3−2k
$$\:\Rightarrow\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\:\mathrm{b}_{\mathrm{n}+\mathrm{1}} +\mathrm{k} \\ $$$$\:\Rightarrow\mathrm{3}−\mathrm{a}_{\mathrm{n}} =\:\mathrm{b}_{\mathrm{n}+\mathrm{1}} +\mathrm{k} \\ $$$$\:\Rightarrow\mathrm{3}−\left(\mathrm{b}_{\mathrm{n}} +\mathrm{k}\right)=\:\mathrm{b}_{\mathrm{n}+\mathrm{1}} +\mathrm{k} \\ $$$$\:\:\Rightarrow\mathrm{b}_{\mathrm{n}+\mathrm{1}} +\mathrm{b}_{\mathrm{n}} =\:\mathrm{3}−\mathrm{2k}\: \\ $$
Commented by mr W last updated on 03/May/24
but your question is ⇒3−6a_n = b_(n+1) +k
$${but}\:{your}\:{question}\:{is} \\ $$$$\:\Rightarrow\mathrm{3}−\mathrm{6a}_{\mathrm{n}} =\:\mathrm{b}_{\mathrm{n}+\mathrm{1}} +\mathrm{k} \\ $$
Commented by efronzo1 last updated on 03/May/24
why 3−7k=0 ?
$$\mathrm{why}\:\mathrm{3}−\mathrm{7k}=\mathrm{0}\:? \\ $$
Commented by mr W last updated on 03/May/24
we just set 3−7k=0 such that b_n becomes an easy G.P.: b_n =−6b_(n−1)
$${we}\:{just}\:{set}\:\mathrm{3}−\mathrm{7}{k}=\mathrm{0}\:{such}\:{that} \\ $$$${b}_{{n}} \:{becomes}\:{an}\:{easy}\:{G}.{P}.: \\ $$$${b}_{{n}} =−\mathrm{6}{b}_{{n}−\mathrm{1}} \\ $$
Commented by mathzup last updated on 03/May/24
the method of sir mrw is correct
$${the}\:{method}\:{of}\:{sir}\:{mrw}\:{is}\:{correct} \\ $$$$ \\ $$
Answered by mathzup last updated on 03/May/24
⇒a_(n+1) +6a_n −3=0 he→r+6=0 ⇒r=−6 ⇒a_n =λ(−6)^n +ρ a_4 =λ(−6)^4 +ρ a_5 =3−6a_4 =3−6(−9)=3+6.9 =57=λ(−6)^5 +ρ we get the system { (((−6)^4 λ+ρ =−9)),(((−6)^5 λ +ρ =57 ⇒)) :} ((−6)^5 −(−6)^4 )λ=57+9 =66 ⇒ λ=((66)/(−6^5 −6^4 ))=−((66)/(6^4 +6^5 ))=−((66)/(6^4 ×7)) ρ=−9−6^4 λ =−9+6^4 .((66)/(6^4 .7)) =−9+((66)/7) rest to finich to calculus...
$$\Rightarrow{a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} −\mathrm{3}=\mathrm{0} \\ $$$${he}\rightarrow{r}+\mathrm{6}=\mathrm{0}\:\Rightarrow{r}=−\mathrm{6}\:\Rightarrow{a}_{{n}} =\lambda\left(−\mathrm{6}\right)^{{n}} \:+\rho \\ $$$${a}_{\mathrm{4}} =\lambda\left(−\mathrm{6}\right)^{\mathrm{4}} +\rho \\ $$$${a}_{\mathrm{5}} =\mathrm{3}−\mathrm{6}{a}_{\mathrm{4}} =\mathrm{3}−\mathrm{6}\left(−\mathrm{9}\right)=\mathrm{3}+\mathrm{6}.\mathrm{9} \\ $$$$=\mathrm{57}=\lambda\left(−\mathrm{6}\right)^{\mathrm{5}} \:+\rho\:\:{we}\:{get}\:{the}\:{system} \\ $$$$\begin{cases}{\left(−\mathrm{6}\right)^{\mathrm{4}} \lambda+\rho\:=−\mathrm{9}}\\{\left(−\mathrm{6}\right)^{\mathrm{5}} \lambda\:+\rho\:=\mathrm{57}\:\:\Rightarrow}\end{cases} \\ $$$$\left(\left(−\mathrm{6}\right)^{\mathrm{5}} −\left(−\mathrm{6}\right)^{\mathrm{4}} \right)\lambda=\mathrm{57}+\mathrm{9}\:=\mathrm{66}\:\Rightarrow \\ $$$$\lambda=\frac{\mathrm{66}}{−\mathrm{6}^{\mathrm{5}} −\mathrm{6}^{\mathrm{4}} }=−\frac{\mathrm{66}}{\mathrm{6}^{\mathrm{4}} +\mathrm{6}^{\mathrm{5}} }=−\frac{\mathrm{66}}{\mathrm{6}^{\mathrm{4}} ×\mathrm{7}} \\ $$$$\rho=−\mathrm{9}−\mathrm{6}^{\mathrm{4}} \:\lambda\:=−\mathrm{9}+\mathrm{6}^{\mathrm{4}} .\frac{\mathrm{66}}{\mathrm{6}^{\mathrm{4}} .\mathrm{7}} \\ $$$$=−\mathrm{9}+\frac{\mathrm{66}}{\mathrm{7}}\:\:{rest}\:{to}\:{finich}\:{to}\:{calculus}… \\ $$

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