# What are first terms of this sequence: f(1)=-2, f(n)=f(n-1)+4?

Jan 8, 2017

$n = 1 \to {a}_{1} = - 2 \leftarrow \text{ given value}$

$n = 2 \to {a}_{2} = - 2 + 4 = 2$
$n = 3 \to {a}_{3} = - 2 + 4 + 4 = 6$
$n = 4 \to {a}_{4} = - 2 + 4 + 4 + 4 = 10$

#### Explanation:

Let the place count be $n$
Let the ${n}^{\text{th}}$ term be ${a}_{n}$

Given $f \left(n = 1\right) = - 2$

We are also told that any one term is the previous term + 4.
This is derived from $f \left(n\right) = f \left(n - 1\right) + 4$ where $f \left(n - 1\right)$ is the previous term.

Consequently we have an Arithmetic sequence with common difference of +4

From this the sequence is:

$n = 1 \to {a}_{1} = - 2 \leftarrow \text{ given value}$

$n = 2 \to {a}_{2} = - 2 + 4 = 2$
$n = 3 \to {a}_{3} = - 2 + 4 + 4 = 6$
$n = 4 \to {a}_{4} = - 2 + 4 + 4 + 4 = 10$

And so on. Also from this we also have an alternative equation for any ${a}_{n}$ in that we have:

${a}_{n} = - 2 + 4 \left(n - 1\right)$