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

Jul 2, 2017

$\left\{0 , 4 , - 12 , 20 , - 44\right\}$

#### Explanation:

We have a sequence:

$\left\{{a}_{n}\right\} = \left\{{a}_{1} , {a}_{2} , {a}_{3} , \ldots\right\}$

Where:

$\setminus \setminus \setminus \setminus {a}_{1} = 0$
${a}_{n + 1} = - 2 {a}_{n} - 4$

Put $n = 1 \implies {a}_{2} = - 2 {a}_{1} - 4$
$\therefore {a}_{2} = - 2 \left(0\right) - 4$
$\text{ } = 0 - 4$
$\text{ } = 4$

Put $n = 2 \implies {a}_{3} = - 2 {a}_{2} - 4$
$\therefore {a}_{3} = - 2 \left(4\right) - 4$
$\text{ } = - 8 - 4$
$\text{ } = - 12$

Put $n = 3 \implies {a}_{4} = - 2 {a}_{3} - 4$
$\therefore {a}_{4} = - 2 \left(- 12\right) - 4$
$\text{ } = 24 - 4$
$\text{ } = 20$

Put $n = 4 \implies {a}_{5} = - 2 {a}_{4} - 4$
$\therefore {a}_{5} = - 2 \left(20\right) - 4$
$\text{ } = - 40 - 4$
$\text{ } = - 44$

Thus the first five terms $\left\{{a}_{1} , {a}_{2} , {a}_{3} , {a}_{5}\right\}$ of the sequence $\left\{{a}_{n}\right\}$ are:

$\left\{0 , 4 , - 12 , 20 , - 44\right\}$