Menu Close

Question-105536




Question Number 105536 by I want to learn more last updated on 29/Jul/20
Answered by adhigenz last updated on 29/Jul/20
α+β = 5, αβ = −2  x_n  = α^n −β^n         = (α+β)(α^(n−1) −β^(n−1) )−αβ(α^(n−2) −β^(n−2) )        = 5x_(n−1) +2x_(n−2)   Substitute for n = 2002:  x_(2002)  = 5x_(2001) +2x_(2000)   ⇒ 2x_(2000) −x_(2002)  = −5x_(2001)   ((2x_(2000) −x_(2002) )/x_(2001) ) = ((−5x_(2001) )/x_(2001) ) = −5
$$\alpha+\beta\:=\:\mathrm{5},\:\alpha\beta\:=\:−\mathrm{2} \\ $$$${x}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}} \\ $$$$\:\:\:\:\:\:=\:\left(\alpha+\beta\right)\left(\alpha^{{n}−\mathrm{1}} −\beta^{{n}−\mathrm{1}} \right)−\alpha\beta\left(\alpha^{{n}−\mathrm{2}} −\beta^{{n}−\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:=\:\mathrm{5}{x}_{{n}−\mathrm{1}} +\mathrm{2}{x}_{{n}−\mathrm{2}} \\ $$$$\mathrm{Substitute}\:\mathrm{for}\:{n}\:=\:\mathrm{2002}: \\ $$$${x}_{\mathrm{2002}} \:=\:\mathrm{5}{x}_{\mathrm{2001}} +\mathrm{2}{x}_{\mathrm{2000}} \\ $$$$\Rightarrow\:\mathrm{2}{x}_{\mathrm{2000}} −{x}_{\mathrm{2002}} \:=\:−\mathrm{5}{x}_{\mathrm{2001}} \\ $$$$\frac{\mathrm{2}{x}_{\mathrm{2000}} −{x}_{\mathrm{2002}} }{{x}_{\mathrm{2001}} }\:=\:\frac{−\mathrm{5}{x}_{\mathrm{2001}} }{{x}_{\mathrm{2001}} }\:=\:−\mathrm{5} \\ $$
Commented by I want to learn more last updated on 29/Jul/20
Wow, thanks sir, but how do we know where to stop in the expansion of    α^n   −  β^n    and how we get the expansion.
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{how}\:\mathrm{do}\:\mathrm{we}\:\mathrm{know}\:\mathrm{where}\:\mathrm{to}\:\mathrm{stop}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\:\:\alpha^{\mathrm{n}} \:\:−\:\:\beta^{\mathrm{n}} \:\:\:\mathrm{and}\:\mathrm{how}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{expansion}. \\ $$
Commented by adhigenz last updated on 30/Jul/20
We can relate on what is on the given  question like α+β and αβ then find the  relationship involving x_n  = α^n −β^n   α^n −β^n  = (α+β)(α^(n−1) −β^(n−1) )+αβ^(n−1) −α^(n−1) β  = (α+β)(α^(n−1) −β^(n−1) )−αβ(α^(n−2) −β^(n−2) )
$$\mathrm{We}\:\mathrm{can}\:\mathrm{relate}\:\mathrm{on}\:\mathrm{what}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{question}\:\mathrm{like}\:\alpha+\beta\:\mathrm{and}\:\alpha\beta\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{relationship}\:\mathrm{involving}\:{x}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}} \\ $$$$\alpha^{{n}} −\beta^{{n}} \:=\:\left(\alpha+\beta\right)\left(\alpha^{{n}−\mathrm{1}} −\beta^{{n}−\mathrm{1}} \right)+\alpha\beta^{{n}−\mathrm{1}} −\alpha^{{n}−\mathrm{1}} \beta \\ $$$$=\:\left(\alpha+\beta\right)\left(\alpha^{{n}−\mathrm{1}} −\beta^{{n}−\mathrm{1}} \right)−\alpha\beta\left(\alpha^{{n}−\mathrm{2}} −\beta^{{n}−\mathrm{2}} \right) \\ $$
Commented by I want to learn more last updated on 30/Jul/20
I appreciate sir
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *