How do you find the first five terms of each sequence $a_{1} = 0$ $a_1=0$ , $a_{n + 1} = - 2 a_{n} - 4$ $a_(n+1)=-2a_n-4$ ?

Question

1790 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-02T22:03:10

${0, 4, - 12, 20, - 44}$ ${0,4,-12,20,-44}$

We have a sequence:

${a_n} = {a_1, a_2, a_3, ... }$

Where:

$\ \ \ \ a_1 =0$
$a_(n+1) = -2a_n-4$

Put $n=1 => a_2 = -2a_1 - 4$
$:. a_2 = -2(0) -4$
$" " = 0-4$
$" " = 4$

Put $n=2 => a_3 = -2a_2 - 4$
$:. a_3 = -2(4) - 4$
$" " = -8-4$
$" " = -12$

Put $n=3 => a_4 = -2a_3 - 4$
$:. a_4 = -2(-12) - 4$
$" " = 24-4$
$" " = 20$

Put $n=4 => a_5 = -2a_4 - 4$
$:. a_5 = -2(20) - 4$
$" " = -40-4$
$" " = -44$

Thus the first five terms ${a_1, a_2, a_3, a_5}$ of the sequence ${a_n}$ are:

${0,4,-12,20,-44}$

How do you find the first five terms of each sequence a_1=0a1=0a_1=0, a_(n+1)=-2a_n-4an+1=−2an−4a_(n+1)=-2a_n-4?