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

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

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Answer 1 · 2016-11-29T17:55:10

Use the definition

Explanation:

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

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

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How do you find the first five terms of each sequence a1=1a_1=1, a2=2a_2=2 an+2=4(an+1)−3ana_(n+2)=4(a_(n+1))-3a_n?

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