Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 105536 by I want to learn more last updated on 29/Jul/20

Answered by adhigenz last updated on 29/Jul/20

$$\alpha +\beta =5,\alpha \beta =-2$$$$x_n={\alpha }^n-{\beta }^n$$$$=(\alpha +\beta )({\alpha }^{n-1}-{\beta }^{n-1})-\alpha \beta ({\alpha }^{n-2}-{\beta }^{n-2})$$$$=5x_{n-1}+2x_{n-2}$$$$\mathrm{Substitute}\mathrm{for}n=2002:$$$$x_{2002}=5x_{2001}+2x_{2000}$$$$\Rightarrow 2x_{2000}-x_{2002}=-5x_{2001}$$$$\frac{2x_{2000}-x_{2002}}{x_{2001}}=\frac{-5x_{2001}}{x_{2001}}=-5$$

Commented by I want to learn more last updated on 29/Jul/20

$$\mathrm{Wow},\mathrm{thanks}\mathrm{sir},\mathrm{but}\mathrm{how}\mathrm{do}\mathrm{we}\mathrm{know}\mathrm{where}\mathrm{to}\mathrm{stop}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}$$$${\alpha }^{\mathrm{n}}-{\beta }^{\mathrm{n}}\mathrm{and}\mathrm{how}\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{expansion}.$$

Commented by adhigenz last updated on 30/Jul/20

$$\mathrm{We}\mathrm{can}\mathrm{relate}\mathrm{on}\mathrm{what}\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{given}$$$$\mathrm{question}\mathrm{like}\alpha +\beta \mathrm{and}\alpha \beta \mathrm{then}\mathrm{find}\mathrm{the}$$$$\mathrm{relationship}\mathrm{involving}{x}_n={\alpha }^n-{\beta }^n$$$${\alpha }^n-{\beta }^n=(\alpha +\beta )({\alpha }^{n-1}-{\beta }^{n-1})+\alpha {\beta }^{n-1}-{\alpha }^{n-1}\beta$$$$=(\alpha +\beta )({\alpha }^{n-1}-{\beta }^{n-1})-\alpha \beta ({\alpha }^{n-2}-{\beta }^{n-2})$$

Commented by I want to learn more last updated on 30/Jul/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com