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Question Number 131155 by EDWIN88 last updated on 02/Feb/21
Given{an+2=an+1+12ana1=3;a2=2findan.
Answered by john_santu last updated on 02/Feb/21
Characteristicequationλ2−λ−12=0λ=1±1+22=1±32soan=C1(1+32)n+C2(1−32)2{a1=3=(1+32)C1+(1−32)C2a2=2=(2+32)C1+(2−32)C2{(1+3)C1+(1−3)C2=6(2+3)C1+(2−3)C2=4(1+31−32+32−3)(C1C2)=(64)C1=|61−342−3|23=8−2323=4−33C2=|1+362+34|23=−8−2323=−4+33⇔an=(4−33)(1+32)n−(4+33)(1−32)n
Answered by mr W last updated on 02/Feb/21
p2−p−12=0p=1±32an=A(1+32)n−1+B(1−32)n−1a1=A+B=3a2=A(1+32)+B(1−32)=2A(1+32)+(3−A)(1−32)=2⇒A=9+36⇒B=3−9+36=9−36⇒an=(9+36)(1+32)n−1+(9−36)(1−32)n−1or⇒an=(43−33)(1+32)n−(43+33)(1−32)n
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