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Question Number 87398 by jagoll last updated on 04/Apr/20

dear mr w a_(n+2) = a_(n+1) − a_n find a_n

$$\mathrm{dear}\:\mathrm{mr}\:\mathrm{w} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:−\:\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \\ $$

Commented by mr W last updated on 04/Apr/20

x^2 −x+1=0 x=((1±i(√3))/2)=cos (π/3)±i sin (π/3)=e^(±i(π/3)) ⇒a_n =Ae^(i((nπ)/3)) +Be^(−i((nπ)/3)) you need to know a_1 and a_2 to determine A and B.

$${x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\pm{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{3}}={e}^{\pm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\Rightarrow{a}_{{n}} ={Ae}^{{i}\frac{{n}\pi}{\mathrm{3}}} +{Be}^{−{i}\frac{{n}\pi}{\mathrm{3}}} \\ $$$${you}\:{need}\:{to}\:{know}\:{a}_{\mathrm{1}} \:{and}\:{a}_{\mathrm{2}} \:{to}\:{determine} \\ $$$${A}\:{and}\:{B}. \\ $$

Commented by jagoll last updated on 04/Apr/20

given Σ_(n = 1) ^(1945) a_n = 2018 and Σ_(n = 1) ^(2018) a_n = 1945. Evaluate? Σ_(n = 1) ^(2001) a_n = ?

$$\mathrm{given}\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\mathrm{1945}} {\sum}}\mathrm{a}_{\mathrm{n}} \:=\:\mathrm{2018}\:\mathrm{and}\: \\ $$$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}\mathrm{a}_{\mathrm{n}} \:=\:\mathrm{1945}.\:\mathrm{Evaluate}? \\ $$$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\mathrm{2001}} {\sum}}\:\mathrm{a}_{\mathrm{n}} \:=\:? \\ $$

Commented by jagoll last updated on 04/Apr/20

how to get A & B sir?

$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{A}\:\&\:\mathrm{B}\:\mathrm{sir}? \\ $$

Commented by mr W last updated on 04/Apr/20

a_n =Ae^(i((nπ)/3)) +Be^(−i((nπ)/3)) =Ap^n +Bq^n Σ_(n=1) ^(1945) a_n =A(Σ_(n=1) ^(1945) p^n )+B(Σ_(n=1) ^(1945) q^n )=2018 ⇒((p(p^(1945) −1))/(p−1))A+((q(q^(1945) −1))/(q−1))B=2018 Σ_(n=1) ^(2018) a_n =A(Σ_(n=1) ^(2018) p^n )+B(Σ_(n=1) ^(2018) q^n )=1945 ⇒((p(p^(2018) −1))/(p−1))A+((q(q^(2018) −1))/(q−1))B=1945 ......

$${a}_{{n}} ={Ae}^{{i}\frac{{n}\pi}{\mathrm{3}}} +{Be}^{−{i}\frac{{n}\pi}{\mathrm{3}}} ={Ap}^{{n}} +{Bq}^{{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{1945}} {\sum}}{a}_{{n}} ={A}\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{1945}} {\sum}}{p}^{{n}} \right)+{B}\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{1945}} {\sum}}{q}^{{n}} \right)=\mathrm{2018} \\ $$$$\Rightarrow\frac{{p}\left({p}^{\mathrm{1945}} −\mathrm{1}\right)}{{p}−\mathrm{1}}{A}+\frac{{q}\left({q}^{\mathrm{1945}} −\mathrm{1}\right)}{{q}−\mathrm{1}}{B}=\mathrm{2018} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}{a}_{{n}} ={A}\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}{p}^{{n}} \right)+{B}\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}{q}^{{n}} \right)=\mathrm{1945} \\ $$$$\Rightarrow\frac{{p}\left({p}^{\mathrm{2018}} −\mathrm{1}\right)}{{p}−\mathrm{1}}{A}+\frac{{q}\left({q}^{\mathrm{2018}} −\mathrm{1}\right)}{{q}−\mathrm{1}}{B}=\mathrm{1945} \\ $$$$...... \\ $$

Commented by jagoll last updated on 04/Apr/20

waw...

$$\mathrm{waw}... \\ $$

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