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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 .

Giventheseries an=3an1+2an2with a1=11&a2=21.Findan.

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

characteristicsequation r23r2=0;r=3±172 an=A.(3+172)n+B(3172)n withn=111=32(A+B)+172(AB) withn=221=132(A+B)+3172(AB) A=(2031717);B=(20+31717) an=(2031717)(3+172)n(20+31717)(3172)n

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