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Question Number 127116 by physicstutes last updated on 27/Dec/20

Given a sequence (u_n ) defined reculsively by u_(n+1) = 3u_n + 4u_(n−1) , u_0 = 1 , u_2 = 3 show that u_(n+1) −4u_n = (−1)^(n+1) (3u_0 −4u_1 ) hence show that u_n is a divegent sequence.

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{sequence}\:\left({u}_{{n}} \right)\:\mathrm{defined}\:\mathrm{reculsively}\:\mathrm{by} \\ $$$$\:{u}_{{n}+\mathrm{1}} \:=\:\mathrm{3}{u}_{{n}} +\:\mathrm{4}{u}_{{n}−\mathrm{1}} ,\:\:{u}_{\mathrm{0}} =\:\mathrm{1}\:,\:{u}_{\mathrm{2}} \:=\:\mathrm{3} \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\:{u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} \:=\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left(\mathrm{3}{u}_{\mathrm{0}} −\mathrm{4}{u}_{\mathrm{1}} \right) \\ $$$$\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\:{u}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{divegent}\:\mathrm{sequence}. \\ $$

Commented by 676597498 last updated on 27/Dec/20

f.maths gce 2020. i got the mark guide

$${f}.{maths}\:{gce}\:\mathrm{2020}. \\ $$$${i}\:{got}\:{the}\:{mark}\:{guide} \\ $$

Commented by Ar Brandon last updated on 27/Dec/20

Quel pays ?

Answered by mr W last updated on 27/Dec/20

p^2 −3p−4=0 (p−4)(p+1)=0 p=4,−1 u_n =A×4^n +B×(−1)^n u_2 =16A+B=3 u_0 =A+B=1 ⇒A=(2/(15)) ⇒B=((13)/(15)) u_n =(2/(15))×4^n +((13)/(15))(−1)^n u_1 =(8/(15))−((13)/(15))=−(1/3) 3u_0 −4u_1 =3+(4/3)=((13)/3)=5B u_(n+1) −4u_n =A×4^(n+1) +B×(−1)^(n+1) −4A×4^n −4B×(−1)^n =5B×(−1)^(n+1) =(−1)^(n+1) (3u_0 −4u_1 )

$${p}^{\mathrm{2}} −\mathrm{3}{p}−\mathrm{4}=\mathrm{0} \\ $$$$\left({p}−\mathrm{4}\right)\left({p}+\mathrm{1}\right)=\mathrm{0} \\ $$$${p}=\mathrm{4},−\mathrm{1} \\ $$$${u}_{{n}} ={A}×\mathrm{4}^{{n}} +{B}×\left(−\mathrm{1}\right)^{{n}} \\ $$$${u}_{\mathrm{2}} =\mathrm{16}{A}+{B}=\mathrm{3} \\ $$$${u}_{\mathrm{0}} ={A}+{B}=\mathrm{1} \\ $$$$\Rightarrow{A}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{13}}{\mathrm{15}} \\ $$$${u}_{{n}} =\frac{\mathrm{2}}{\mathrm{15}}×\mathrm{4}^{{n}} +\frac{\mathrm{13}}{\mathrm{15}}\left(−\mathrm{1}\right)^{{n}} \\ $$$${u}_{\mathrm{1}} =\frac{\mathrm{8}}{\mathrm{15}}−\frac{\mathrm{13}}{\mathrm{15}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{3}{u}_{\mathrm{0}} −\mathrm{4}{u}_{\mathrm{1}} =\mathrm{3}+\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{13}}{\mathrm{3}}=\mathrm{5}{B} \\ $$$${u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} ={A}×\mathrm{4}^{{n}+\mathrm{1}} +{B}×\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} −\mathrm{4}{A}×\mathrm{4}^{{n}} −\mathrm{4}{B}×\left(−\mathrm{1}\right)^{{n}} \\ $$$$=\mathrm{5}{B}×\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} =\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left(\mathrm{3}{u}_{\mathrm{0}} −\mathrm{4}{u}_{\mathrm{1}} \right) \\ $$

Commented by physicstutes last updated on 27/Dec/20

thats wonderful gracias

$$\mathrm{thats}\:\mathrm{wonderful}\:\mathrm{gracias} \\ $$

Answered by Raxreedoroid last updated on 27/Dec/20

Given a sequence (u_n ) defined reculsively by u_(n+1) = 3u_n + 4u_(n−1) , u_0 = 1 , u_2 = 3 show that u_(n+1) −4u_n = (−1)^(n+1) (3u_0 −4u_1 ) hence show that u_n is a divegent sequence. 3u_1 =u_2 −4u_0 u_1 =((3−4)/3)=−0.3^_ u_(n+1) −4u_n =(−1)^(n+1) (3×1+4×0.3^_ ) u_(n+1) −4u_n =4.3^− (−1)^(n+1) Let n=1 then u_2 −4u_1 =(−1)^(1+1) (4.3^_ ) 3−4(−0.3^− )=4.3^− 4.3^− =4.3^− ✓ (a):Assume u_(n+1) −4u_n =4.3^− (−1)^(n+1) Subtitute x as 4.3^− (−1)^(n+1 ) u_(n+1) −4u_n =x solve for x u_(n+1) =3u_n +4u_(n−1) ∴ 3u_n +4u_(n−1) −4u_n =x x=4u_(n−1) −u_n ∴ −x=u_n −4u_(n−1) u_n −4u_(n−1) =4.3^− (−1)(−1)^(n+1) u_n −4u_(n−1) =4.3^− (−1)^(n+2) =4.3^− (−1)^2 (−1)^n =4.3^− (−1)^n Subsitute n as n+1 u_(n+1) −4u_n =4.3^− (−1)^(n+1) which is true from (a) u_(n+1) =4.3^− (−1)^(n+1) +4u_n =4.3^− (−1)^(n+1) +4×4.3^− (−1)^n +16u_(n−1) ...4^n (−1)^(n+1) +4^(n+1) u_(n+1) =4^(n+1) +Σ_(i=1) ^(n+1) 4^(i−1) ×4.3^− (−1)^(n−i) u_n =4^n +4.3^− Σ_(i=1) ^n 4^(i−1) (−1)^(n−i−1) u_n =4^n +4.3^− Σ_(i=1) ^n 4^(i−1) (−1)^(n−i−1) l_(n→∞) im (4^n +4.3^− Σ_(i=1) ^n 4^(i−1) (−1)^(n−i−1) )=∞

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{sequence}\:\left({u}_{{n}} \right)\:\mathrm{defined}\:\mathrm{reculsively}\:\mathrm{by} \\ $$$$\:{u}_{{n}+\mathrm{1}} \:=\:\mathrm{3}{u}_{{n}} +\:\mathrm{4}{u}_{{n}−\mathrm{1}} ,\:\:{u}_{\mathrm{0}} =\:\mathrm{1}\:,\:{u}_{\mathrm{2}} \:=\:\mathrm{3} \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\:{u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} \:=\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left(\mathrm{3}{u}_{\mathrm{0}} −\mathrm{4}{u}_{\mathrm{1}} \right) \\ $$$$\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\:{u}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{divegent}\:\mathrm{sequence}. \\ $$$$\mathrm{3}{u}_{\mathrm{1}} ={u}_{\mathrm{2}} −\mathrm{4}{u}_{\mathrm{0}} \\ $$$${u}_{\mathrm{1}} =\frac{\mathrm{3}−\mathrm{4}}{\mathrm{3}}=−\mathrm{0}.\overset{\_} {\mathrm{3}} \\ $$$${u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left(\mathrm{3}×\mathrm{1}+\mathrm{4}×\mathrm{0}.\overset{\_} {\mathrm{3}}\right) \\ $$$${u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${Let}\:{n}=\mathrm{1}\:{then}\:{u}_{\mathrm{2}} −\mathrm{4}{u}_{\mathrm{1}} =\left(−\mathrm{1}\right)^{\mathrm{1}+\mathrm{1}} \left(\mathrm{4}.\overset{\_} {\mathrm{3}}\right) \\ $$$$\mathrm{3}−\mathrm{4}\left(−\mathrm{0}.\overset{−} {\mathrm{3}}\right)=\mathrm{4}.\overset{−} {\mathrm{3}} \\ $$$$\mathrm{4}.\overset{−} {\mathrm{3}}=\mathrm{4}.\overset{−} {\mathrm{3}}\checkmark \\ $$$$\left({a}\right):{Assume}\:{u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${Subtitute}\:{x}\:{as}\:\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}\:\:\:\:} \:\:\:{u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} ={x} \\ $$$${solve}\:{for}\:{x} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{3}{u}_{{n}} +\mathrm{4}{u}_{{n}−\mathrm{1}} \:\:\:\therefore\:\mathrm{3}{u}_{{n}} +\mathrm{4}{u}_{{n}−\mathrm{1}} −\mathrm{4}{u}_{{n}} ={x} \\ $$$${x}=\mathrm{4}{u}_{{n}−\mathrm{1}} −{u}_{{n}} \\ $$$$\therefore\:\:−{x}={u}_{{n}} −\mathrm{4}{u}_{{n}−\mathrm{1}} \\ $$$${u}_{{n}} −\mathrm{4}{u}_{{n}−\mathrm{1}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${u}_{{n}} −\mathrm{4}{u}_{{n}−\mathrm{1}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{2}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}} \\ $$$${Subsitute}\:{n}\:{as}\:{n}+\mathrm{1} \\ $$$${u}_{{n}+\mathrm{1}} −\mathrm{4}{u}_{{n}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {which}\:{is}\:{true}\:{from}\:\left({a}\right) \\ $$$$\:{u}_{{n}+\mathrm{1}} =\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} +\mathrm{4}{u}_{{n}} \\ $$$$\:=\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} +\mathrm{4}×\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}} +\mathrm{16}{u}_{{n}−\mathrm{1}} ...\mathrm{4}^{{n}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} +\mathrm{4}^{{n}+\mathrm{1}} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{4}^{{n}+\mathrm{1}} +\underset{{i}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\mathrm{4}^{{i}−\mathrm{1}} ×\mathrm{4}.\overset{−} {\mathrm{3}}\left(−\mathrm{1}\right)^{{n}−{i}} \\ $$$${u}_{{n}} =\mathrm{4}^{{n}} +\mathrm{4}.\overset{−} {\mathrm{3}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{4}^{{i}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}−{i}−\mathrm{1}} \\ $$$${u}_{{n}} =\mathrm{4}^{{n}} +\mathrm{4}.\overset{−} {\mathrm{3}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{4}^{{i}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}−{i}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {{l}im}\:\left(\mathrm{4}^{{n}} +\mathrm{4}.\overset{−} {\mathrm{3}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{4}^{{i}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}−{i}−\mathrm{1}} \right)=\infty \\ $$

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