Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 137899 by bemath last updated on 08/Apr/21

Given the series   a_n = 3a_(n−1) + 2a_(n−2)  with   a_1 =11 & a_2  = 21. Find a_n .

$${Given}\:{the}\:{series}\: \\ $$ $${a}_{{n}} =\:\mathrm{3}{a}_{{n}−\mathrm{1}} +\:\mathrm{2}{a}_{{n}−\mathrm{2}} \:{with}\: \\ $$ $${a}_{\mathrm{1}} =\mathrm{11}\:\&\:{a}_{\mathrm{2}} \:=\:\mathrm{21}.\:{Find}\:{a}_{{n}} . \\ $$

Answered by benjo_mathlover last updated on 08/Apr/21

characteristics equation  r^2 −3r−2 = 0 ; r=((3±(√(17)))/2)  a_n  = A.(((3+(√(17)))/2))^n +B(((3−(√(17)))/2))^n   with n=1⇒11 = (3/2)(A+B)+((√(17))/2)(A−B)  with n=2⇒21=((13)/2)(A+B)+((3(√(17)))/2)(A−B)  A=(((20−3(√(17)))/( (√(17)) ))) ; B=−(((20+3(√(17)))/( (√(17)))))  ∴ a_n = (((20−3(√(17)))/( (√(17)))))(((3+(√(17)))/2))^n −(((20+3(√(17)))/( (√(17)))))(((3−(√(17)))/2))^n

$${characteristics}\:{equation} \\ $$ $${r}^{\mathrm{2}} −\mathrm{3}{r}−\mathrm{2}\:=\:\mathrm{0}\:;\:{r}=\frac{\mathrm{3}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$ $${a}_{{n}} \:=\:{A}.\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{{n}} +{B}\left(\frac{\mathrm{3}−\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{{n}} \\ $$ $${with}\:{n}=\mathrm{1}\Rightarrow\mathrm{11}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\left({A}+{B}\right)+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\left({A}−{B}\right) \\ $$ $${with}\:{n}=\mathrm{2}\Rightarrow\mathrm{21}=\frac{\mathrm{13}}{\mathrm{2}}\left({A}+{B}\right)+\frac{\mathrm{3}\sqrt{\mathrm{17}}}{\mathrm{2}}\left({A}−{B}\right) \\ $$ $${A}=\left(\frac{\mathrm{20}−\mathrm{3}\sqrt{\mathrm{17}}}{\:\sqrt{\mathrm{17}}\:}\right)\:;\:{B}=−\left(\frac{\mathrm{20}+\mathrm{3}\sqrt{\mathrm{17}}}{\:\sqrt{\mathrm{17}}}\right) \\ $$ $$\therefore\:{a}_{{n}} =\:\left(\frac{\mathrm{20}−\mathrm{3}\sqrt{\mathrm{17}}}{\:\sqrt{\mathrm{17}}}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{20}+\mathrm{3}\sqrt{\mathrm{17}}}{\:\sqrt{\mathrm{17}}}\right)\left(\frac{\mathrm{3}−\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com