How do you find the first five terms of each sequence $a_1=1$ , $a_2=2$ $a_(n+2)=4(a_(n+1))-3a_n$ ?

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Answer 1 · 2016-11-29T17:55:10

Use the definition

$a_1=1$
$a_2=2$
$n=1: a_3=4a_2-3a_1 =8-3=5$
$n=2: a_4=4a_3-3a_2 =20-6=14$
$n=3: a_5=4a_4-3a_3 =56-15=41$

How do you find the first five terms of each sequence a_1=1a_1=1, a_2=2a_2=2 a_(n+2)=4(a_(n+1))-3a_na_(n+2)=4(a_(n+1))-3a_n?