How do you find the first four terms given $a_1 = 5$ ; $a_n = 2a_(n-1) + 4$ ?

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Answer 1 · 2016-08-02T23:01:20

$5,14,32,68$

${ (a_1 = 5), (a_n = 2a_(n-1)+4) :}$

This formulation describes the sequence $a_1, a_2, a_3,...$ recursively.

We find:

$a_1 = 5$

$a_2 = 2a_1+4 = 2(5)+4 = 14$

$a_3 = 2a_2+4 = 2(14)+4 = 32$

$a_4 = 2a_3+4 = 2(32)+4 = 68$

$color(white)()$
Bonus

The general formula for a term of this sequence is described by the formula:

$a_n = 9*2^(n-1)-4$

as we can verify with a proof by induction:

Base case:

$9*2^((1)-1) - 4 = 9*2^0-4 = 9*1-4 = 5 = a_1$

Induction step:

$9*2^((n)-1) - 4 = 2(9*2^((n-1)-1)-4)+4 = 2a_(n-1)+4$

How do you find the first four terms given a_1 = 5a_1 = 5; a_n = 2a_(n-1) + 4a_n = 2a_(n-1) + 4?