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Question Number 75616 by mr W last updated on 13/Dec/19

A_1 =2 B_1 =1 A_(n+1) =A_n −8B_n B_(n+1) =A_n +7B_n find A_n =?, B_n =?

$${A}_{\mathrm{1}} =\mathrm{2} \\ $$$${B}_{\mathrm{1}} =\mathrm{1} \\ $$$${A}_{{n}+\mathrm{1}} ={A}_{{n}} −\mathrm{8}{B}_{{n}} \\ $$$${B}_{{n}+\mathrm{1}} ={A}_{{n}} +\mathrm{7}{B}_{{n}} \\ $$$${find}\:{A}_{{n}} =?,\:{B}_{{n}} =? \\ $$

Commented by mind is power last updated on 13/Dec/19

((A_(n+1) ),(B_(n+1) ) )= (((1 −8)),((1 7)) ) ((A_n ),(B_n ) ) M= (((1 −8)),((1 7)) ) (((1 −x −8)),((1 7−x)) )=0⇒x^2 −8x+15=0 x=5,x=3 D= (((3 0)),((0 5)) ) x=3⇒−2x−8y=0⇒x=−4y U=(−4,1) x=5⇒x+2y=0⇒x=−2y⇒V=(−2,1) P= (((−4 −2)),((1 1)) ) p^− =−(1/2) (((1 2)),((−1 −4)) ) M=P D P^− M^n =PD^n P^− ((A_n ),(B_n ) )=PD^(n−1) .P^− . ((A_1 ),(B_1 ) ) PD^n P^− =P. (((3^n 0)),((0 5^n )) ).−(1/2) (((1 2)),((−1 −4)) ) = (((−4 −2)),((1 1)) ) (((−(3^n /2) −3^n )),(((5^n /2) 2.5^n )) )= (((2.3^n −5^n 4.3^n −4.5^n )),((−(3^n /2)+(5^n /2) −3^n +2.5^n )) ) ((A_(n+1) ),(B_(n+1) ) )= (((2.3^n −5^n 4.3^n −4.5^n )),((−(3^n /2)+(5^n /2) −3^n +2.5^n )) ) ((2),(1) ) A_(n+1) =8.3^n −6.5^n B_(n+1) =−2.3^n +3.5^n A_n =8.3^(n−1) −6.5^(n−1) B_n =−2.3^(n−1) +3.5^(n−1)

$$\begin{pmatrix}{\mathrm{A}_{\mathrm{n}+\mathrm{1}} }\\{\mathrm{B}_{\mathrm{n}+\mathrm{1}} }\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:−\mathrm{8}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{7}}\end{pmatrix}\begin{pmatrix}{\mathrm{A}_{\mathrm{n}} }\\{\mathrm{B}_{\mathrm{n}} }\end{pmatrix} \\ $$$$\mathrm{M}=\begin{pmatrix}{\mathrm{1}\:\:\:\:−\mathrm{8}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{7}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{1}\:−\mathrm{x}\:\:\:−\mathrm{8}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{7}−\mathrm{x}}\end{pmatrix}=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{5},\mathrm{x}=\mathrm{3} \\ $$$$\mathrm{D}=\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{5}}\end{pmatrix} \\ $$$$\mathrm{x}=\mathrm{3}\Rightarrow−\mathrm{2x}−\mathrm{8y}=\mathrm{0}\Rightarrow\mathrm{x}=−\mathrm{4y} \\ $$$$\mathrm{U}=\left(−\mathrm{4},\mathrm{1}\right) \\ $$$$\mathrm{x}=\mathrm{5}\Rightarrow\mathrm{x}+\mathrm{2y}=\mathrm{0}\Rightarrow\mathrm{x}=−\mathrm{2y}\Rightarrow\mathrm{V}=\left(−\mathrm{2},\mathrm{1}\right) \\ $$$$\mathrm{P}=\begin{pmatrix}{−\mathrm{4}\:\:\:\:\:\:\:−\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\: \\ $$$$\mathrm{p}^{−} =−\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$$\mathrm{M}=\mathrm{P}\:\:\mathrm{D}\:\:\mathrm{P}^{−} \\ $$$$\mathrm{M}^{\mathrm{n}} =\mathrm{PD}^{\mathrm{n}} \mathrm{P}^{−} \\ $$$$\begin{pmatrix}{\mathrm{A}_{\mathrm{n}} }\\{\mathrm{B}_{\mathrm{n}} }\end{pmatrix}=\mathrm{PD}^{\mathrm{n}−\mathrm{1}} .\mathrm{P}^{−} .\begin{pmatrix}{\mathrm{A}_{\mathrm{1}} }\\{\mathrm{B}_{\mathrm{1}} }\end{pmatrix} \\ $$$$\mathrm{PD}^{\mathrm{n}} \mathrm{P}^{−} =\mathrm{P}.\begin{pmatrix}{\mathrm{3}^{\mathrm{n}} \:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{n}} }\end{pmatrix}.−\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{−\mathrm{4}\:\:\:\:\:\:\:\:\:−\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{−\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:−\mathrm{3}^{\mathrm{n}} \:\:}\\{\frac{\mathrm{5}^{\mathrm{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\mathrm{5}^{\mathrm{n}} }\end{pmatrix}=\begin{pmatrix}{\mathrm{2}.\mathrm{3}^{\mathrm{n}} −\mathrm{5}^{\mathrm{n}} \:\:\:\:\:\:\:\:\:\mathrm{4}.\mathrm{3}^{\mathrm{n}} −\mathrm{4}.\mathrm{5}^{\mathrm{n}} }\\{−\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{2}}+\frac{\mathrm{5}^{\mathrm{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:−\mathrm{3}^{\mathrm{n}} +\mathrm{2}.\mathrm{5}^{\mathrm{n}} }\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{A}_{\mathrm{n}+\mathrm{1}} }\\{\mathrm{B}_{\mathrm{n}+\mathrm{1}} }\end{pmatrix}=\begin{pmatrix}{\mathrm{2}.\mathrm{3}^{\mathrm{n}} −\mathrm{5}^{\mathrm{n}} \:\:\:\:\:\:\:\mathrm{4}.\mathrm{3}^{\mathrm{n}} −\mathrm{4}.\mathrm{5}^{\mathrm{n}} }\\{−\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{2}}+\frac{\mathrm{5}^{\mathrm{n}} }{\mathrm{2}}\:\:\:\:\:\:\:−\mathrm{3}^{\mathrm{n}} +\mathrm{2}.\mathrm{5}^{\mathrm{n}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{A}_{\mathrm{n}+\mathrm{1}} =\mathrm{8}.\mathrm{3}^{\mathrm{n}} −\mathrm{6}.\mathrm{5}^{\mathrm{n}} \\ $$$$\mathrm{B}_{\mathrm{n}+\mathrm{1}} =−\mathrm{2}.\mathrm{3}^{\mathrm{n}} +\mathrm{3}.\mathrm{5}^{\mathrm{n}} \\ $$$$\mathrm{A}_{\mathrm{n}} =\mathrm{8}.\mathrm{3}^{\mathrm{n}−\mathrm{1}} −\mathrm{6}.\mathrm{5}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{B}_{\mathrm{n}} =−\mathrm{2}.\mathrm{3}^{\mathrm{n}−\mathrm{1}} +\mathrm{3}.\mathrm{5}^{\mathrm{n}−\mathrm{1}} \\ $$$$ \\ $$

Commented by mr W last updated on 14/Dec/19

thanks alot sir! i need some time to understand your solution and why you did so.

$${thanks}\:{alot}\:{sir}! \\ $$$${i}\:{need}\:{some}\:{time}\:{to}\:{understand} \\ $$$${your}\:{solution}\:{and}\:{why}\:{you}\:{did}\:{so}. \\ $$

Commented by peter frank last updated on 14/Dec/19

help me Qn 75602

$${help}\:{me}\:{Qn}\:\mathrm{75602} \\ $$

Commented by mind is power last updated on 14/Dec/19

Hello Sir Mr W , its solve system of series usign lineair Algebre not sur in french Tap in google ∴ System de suite numerique et diagonilisations Des matrice ∵ you get in french but you can understand and my bee you will find after that in english sorry my english is not good

$$\mathrm{Hello}\:\mathrm{Sir}\:\mathrm{Mr}\:\mathrm{W}\:,\:\mathrm{its}\:\mathrm{solve}\:\mathrm{system}\:\mathrm{of}\:\mathrm{series}\:\mathrm{usign}\:\mathrm{lineair}\:\mathrm{Algebre} \\ $$$$\mathrm{not}\:\mathrm{sur}\:\mathrm{in}\:\mathrm{french}\:\:\mathrm{Tap}\:\mathrm{in}\:\mathrm{google}\:\: \\ $$$$\therefore\:\mathrm{System}\:\mathrm{de}\:\mathrm{suite}\:\mathrm{numerique}\:\mathrm{et}\:\mathrm{diagonilisations}\:\mathrm{Des}\:\mathrm{matrice}\:\because \\ $$$$\mathrm{you}\:\mathrm{get}\:\mathrm{in}\:\mathrm{french}\:\mathrm{but}\:\mathrm{you}\:\mathrm{can}\:\mathrm{understand}\:\mathrm{and}\:\mathrm{my}\:\mathrm{bee}\:\mathrm{you}\:\mathrm{will} \\ $$$$\mathrm{find}\:\mathrm{after}\:\mathrm{that}\:\mathrm{in}\:\mathrm{english}\:\mathrm{sorry}\:\mathrm{my}\:\mathrm{english}\:\mathrm{is}\:\mathrm{not}\:\mathrm{good} \\ $$

Commented by mr W last updated on 14/Dec/19

M^(ERCI) B^(EAUCOUP) !

$$\mathbb{M}^{\mathbb{ERCI}} \:\mathbb{B}^{\mathbb{EAUCOUP}} ! \\ $$

Commented by mind is power last updated on 14/Dec/19

De Rien :)

$$\left.\mathrm{De}\:\mathrm{Rien}\::\right) \\ $$

Commented by mathmax by abdo last updated on 15/Dec/19

we have a_(n+1) =a_n −8b_n b_(n+1) =a_n +7b_n ⇒ ((a_(n+1) ),(b_(n+1) ) )= (((1 −8)),((1 7)) ) ((a_n ),(b_n ) ) ⇒ ((a_(n+1) ),(b_(n+1) ) ) =A^n ((a_1 ),(b_1 ) ) let calculate A^n p_c (x)=det(A−xI) = determinant (((1−x −8)),((1 7−x)))=(1−x)(7−x)+8 =7−x−7x+x^2 +8 =x^2 −8x +15 Δ^′ =4^2 −15 =1 ⇒x_1 =4+1 =5 and x_2 =4−1 =3 we have x^n =q p_c (x)+u_n x +v_n ⇒ A^n =u_n A +v_n I we have 3^n = 3u_n +v_n and 5^n =5u_n +v_n ⇒ 2u_n =5^n −3^n ⇒u_n =((5^n −3^n )/2) also v_n =3^n −3u_n =3^n −3×((5^n −3^n )/2) =((2.3^n −3.5^n +3.3^n )/2) =((5.3^n −3.5^n )/2) ⇒ A^n =((5^n −3^n )/2) (((1 −8)),((1 7)) )+ ((5.3^n −3.5^n )/2) (((1 0)),((0 1)) ) = (((((5^n −3^n )/2) −4.5^n +4.3^n )),((((5^n −3^n )/2) (7/2)(5^n −3^n ))) ) + (((((5.3^n −3.5^n )/2) 0)),((0 ((5.3^n −3.5^n )/2))) ) = (((−5^n +2.3^n 4.3^n −4.5^n )),((((5^n −3^n )/2) 2. 5^n −3^n )) ) ⇒ ((a_(n+1) ),(b_(n+1) ) ) =A^n ((2),(1) ) and A^n is known ...

$${we}\:{have}\:\:{a}_{{n}+\mathrm{1}} ={a}_{{n}} −\mathrm{8}{b}_{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{7}{b}_{{n}} \:\:\:\Rightarrow\begin{pmatrix}{{a}_{{n}+\mathrm{1}} }\\{{b}_{{n}+\mathrm{1}} }\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{8}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}}\end{pmatrix}\:\begin{pmatrix}{{a}_{{n}} }\\{{b}_{{n}} }\end{pmatrix} \\ $$$$\Rightarrow\:\begin{pmatrix}{{a}_{{n}+\mathrm{1}} }\\{{b}_{{n}+\mathrm{1}} }\end{pmatrix}\:\:={A}^{{n}} \:\begin{pmatrix}{{a}_{\mathrm{1}} }\\{{b}_{\mathrm{1}} }\end{pmatrix}\:\:\:{let}\:{calculate}\:{A}^{{n}} \\ $$$${p}_{{c}} \left({x}\right)={det}\left({A}−{xI}\right)\:=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:−\mathrm{8}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}−{x}}\end{vmatrix}=\left(\mathrm{1}−{x}\right)\left(\mathrm{7}−{x}\right)+\mathrm{8} \\ $$$$=\mathrm{7}−{x}−\mathrm{7}{x}+{x}^{\mathrm{2}} +\mathrm{8}\:={x}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{15} \\ $$$$\Delta^{'} =\mathrm{4}^{\mathrm{2}} −\mathrm{15}\:=\mathrm{1}\:\Rightarrow{x}_{\mathrm{1}} =\mathrm{4}+\mathrm{1}\:=\mathrm{5}\:\:{and}\:{x}_{\mathrm{2}} =\mathrm{4}−\mathrm{1}\:=\mathrm{3} \\ $$$${we}\:{have}\:{x}^{{n}} ={q}\:{p}_{{c}} \left({x}\right)+{u}_{{n}} {x}\:+{v}_{{n}} \:\Rightarrow\:{A}^{{n}} ={u}_{{n}} {A}\:+{v}_{{n}} {I} \\ $$$${we}\:{have}\:\mathrm{3}^{{n}} =\:\mathrm{3}{u}_{{n}} +{v}_{{n}} \:{and}\:\mathrm{5}^{{n}} \:=\mathrm{5}{u}_{{n}} +{v}_{{n}} \:\Rightarrow \\ $$$$\mathrm{2}{u}_{{n}} =\mathrm{5}^{{n}} −\mathrm{3}^{{n}} \:\Rightarrow{u}_{{n}} =\frac{\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }{\mathrm{2}}\:\:{also}\:{v}_{{n}} =\mathrm{3}^{{n}} −\mathrm{3}{u}_{{n}} =\mathrm{3}^{{n}} −\mathrm{3}×\frac{\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}.\mathrm{3}^{{n}} −\mathrm{3}.\mathrm{5}^{{n}} +\mathrm{3}.\mathrm{3}^{{n}} }{\mathrm{2}}\:=\frac{\mathrm{5}.\mathrm{3}^{{n}} −\mathrm{3}.\mathrm{5}^{{n}} }{\mathrm{2}}\:\Rightarrow \\ $$$${A}^{{n}} =\frac{\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{8}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}}\end{pmatrix}+\:\frac{\mathrm{5}.\mathrm{3}^{{n}} −\mathrm{3}.\mathrm{5}^{{n}} }{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{4}.\mathrm{5}^{{n}} +\mathrm{4}.\mathrm{3}^{{n}} }\\{\frac{\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{7}}{\mathrm{2}}\left(\mathrm{5}^{{n}} −\mathrm{3}^{{n}} \right)}\end{pmatrix}\:\:+\begin{pmatrix}{\frac{\mathrm{5}.\mathrm{3}^{{n}} −\mathrm{3}.\mathrm{5}^{{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}.\mathrm{3}^{{n}} −\mathrm{3}.\mathrm{5}^{{n}} }{\mathrm{2}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{−\mathrm{5}^{{n}} +\mathrm{2}.\mathrm{3}^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}.\mathrm{3}^{{n}} −\mathrm{4}.\mathrm{5}^{{n}} }\\{\frac{\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\:\:\mathrm{5}^{{n}} −\mathrm{3}^{{n}} }\end{pmatrix}\:\Rightarrow \\ $$$$\begin{pmatrix}{{a}_{{n}+\mathrm{1}} }\\{{b}_{{n}+\mathrm{1}} }\end{pmatrix}\:\:={A}^{{n}} \:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:\:\:\:{and}\:{A}^{{n}} \:{is}\:{known}\:... \\ $$

Answered by mr W last updated on 14/Dec/19

attempt an other way: let C_(n+1) =A_(n+1) +kB_(n+1) with k=constant ⇒C_n =A_n +kB_n if we can get C_(n+1) =hC_n with h=constant then we can solve both A_n and B_n . C_(n+1) =A_(n+1) +kB_(n+1) C_(n+1) =(A_n −8B_n )+k(A_n +7B_n ) C_(n+1) =(1+k)A_n +(7k−8)B_n C_(n+1) =(1+k)[A_n +((7k−8)/(1+k))B_n ] ≡C_(n+1) =h[A_n +kB_n ]=hC_n i.e. ((7k−8)/(1+k))=k ⇒k^2 −6k+8=0 ⇒(k−2)(k−4)=0 ⇒k=2 or k=4 h=1+k=3 or 5 with k=2 and h=3: C_1 =A_1 +kB_1 =2+2×1=4 C_(n+1) =hC_n =h^n C_1 =4×3^n i.e. A_(n+1) +2B_(n+1) =4×3^n ...(i) with k=4 and h=5: C_1 =A_1 +kB_1 =2+4×1=6 C_(n+1) =hC_n =h^n C_1 =6×5^n i.e. A_(n+1) +4B_(n+1) =6×5^n ...(ii) (ii)−(i): 2B_(n+1) =6×5^n −4×3^n ⇒B_(n+1) =3×5^n −2×3^n (i)×2−(ii): A_(n+1) =2×4×3^n −6×5^n ⇒A_(n+1) =8×3^n −6×5^n ⇒A_n =8×3^(n−1) −6×5^(n−1) ⇒B_n =3×5^(n−1) −2×3^(n−1)

$${attempt}\:{an}\:{other}\:{way}: \\ $$$${let}\:{C}_{{n}+\mathrm{1}} ={A}_{{n}+\mathrm{1}} +{kB}_{{n}+\mathrm{1}} \:{with}\:{k}={constant} \\ $$$$\Rightarrow{C}_{{n}} ={A}_{{n}} +{kB}_{{n}} \\ $$$${if}\:{we}\:{can}\:{get}\:{C}_{{n}+\mathrm{1}} ={hC}_{{n}} \:{with}\:{h}={constant} \\ $$$${then}\:{we}\:{can}\:{solve}\:{both}\:{A}_{{n}} \:{and}\:{B}_{{n}} . \\ $$$${C}_{{n}+\mathrm{1}} ={A}_{{n}+\mathrm{1}} +{kB}_{{n}+\mathrm{1}} \\ $$$${C}_{{n}+\mathrm{1}} =\left({A}_{{n}} −\mathrm{8}{B}_{{n}} \right)+{k}\left({A}_{{n}} +\mathrm{7}{B}_{{n}} \right) \\ $$$${C}_{{n}+\mathrm{1}} =\left(\mathrm{1}+{k}\right){A}_{{n}} +\left(\mathrm{7}{k}−\mathrm{8}\right){B}_{{n}} \\ $$$${C}_{{n}+\mathrm{1}} =\left(\mathrm{1}+{k}\right)\left[{A}_{{n}} +\frac{\mathrm{7}{k}−\mathrm{8}}{\mathrm{1}+{k}}{B}_{{n}} \right] \\ $$$$\equiv{C}_{{n}+\mathrm{1}} ={h}\left[{A}_{{n}} +{kB}_{{n}} \right]={hC}_{{n}} \\ $$$${i}.{e}.\: \\ $$$$\frac{\mathrm{7}{k}−\mathrm{8}}{\mathrm{1}+{k}}={k} \\ $$$$\Rightarrow{k}^{\mathrm{2}} −\mathrm{6}{k}+\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow\left({k}−\mathrm{2}\right)\left({k}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{2}\:{or}\:{k}=\mathrm{4} \\ $$$${h}=\mathrm{1}+{k}=\mathrm{3}\:{or}\:\mathrm{5} \\ $$$$ \\ $$$${with}\:{k}=\mathrm{2}\:{and}\:{h}=\mathrm{3}: \\ $$$${C}_{\mathrm{1}} ={A}_{\mathrm{1}} +{kB}_{\mathrm{1}} =\mathrm{2}+\mathrm{2}×\mathrm{1}=\mathrm{4} \\ $$$${C}_{{n}+\mathrm{1}} ={hC}_{{n}} ={h}^{{n}} {C}_{\mathrm{1}} =\mathrm{4}×\mathrm{3}^{{n}} \\ $$$${i}.{e}.\:{A}_{{n}+\mathrm{1}} +\mathrm{2}{B}_{{n}+\mathrm{1}} =\mathrm{4}×\mathrm{3}^{{n}} \:\:\:...\left({i}\right) \\ $$$${with}\:{k}=\mathrm{4}\:{and}\:{h}=\mathrm{5}: \\ $$$${C}_{\mathrm{1}} ={A}_{\mathrm{1}} +{kB}_{\mathrm{1}} =\mathrm{2}+\mathrm{4}×\mathrm{1}=\mathrm{6} \\ $$$${C}_{{n}+\mathrm{1}} ={hC}_{{n}} ={h}^{{n}} {C}_{\mathrm{1}} =\mathrm{6}×\mathrm{5}^{{n}} \\ $$$${i}.{e}.\:{A}_{{n}+\mathrm{1}} +\mathrm{4}{B}_{{n}+\mathrm{1}} =\mathrm{6}×\mathrm{5}^{{n}} \:\:\:...\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\mathrm{2}{B}_{{n}+\mathrm{1}} =\mathrm{6}×\mathrm{5}^{{n}} −\mathrm{4}×\mathrm{3}^{{n}} \\ $$$$\Rightarrow{B}_{{n}+\mathrm{1}} =\mathrm{3}×\mathrm{5}^{{n}} −\mathrm{2}×\mathrm{3}^{{n}} \\ $$$$\left({i}\right)×\mathrm{2}−\left({ii}\right): \\ $$$${A}_{{n}+\mathrm{1}} =\mathrm{2}×\mathrm{4}×\mathrm{3}^{{n}} −\mathrm{6}×\mathrm{5}^{{n}} \\ $$$$\Rightarrow{A}_{{n}+\mathrm{1}} =\mathrm{8}×\mathrm{3}^{{n}} −\mathrm{6}×\mathrm{5}^{{n}} \\ $$$$ \\ $$$$\Rightarrow{A}_{{n}} =\mathrm{8}×\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{6}×\mathrm{5}^{{n}−\mathrm{1}} \\ $$$$\Rightarrow{B}_{{n}} =\mathrm{3}×\mathrm{5}^{{n}−\mathrm{1}} −\mathrm{2}×\mathrm{3}^{{n}−\mathrm{1}} \\ $$

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