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

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

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Answer 1 · 2017-01-08T20:12:35

$n = 1 \to a_{1} = - 2 \leftarrow given value$ $n=1->a_1=-2 larr" given value"$

$n = 2 \to a_{2} = - 2 + 4 = 2$ $n=2->a_2=-2+4 = 2$
$n = 3 \to a_{3} = - 2 + 4 + 4 = 6$ $n=3->a_3=-2+4+4 = 6$
$n = 4 \to a_{4} = - 2 + 4 + 4 + 4 = 10$ $n=4->a_4=-2+4+4+4=10$

Explanation:

Let the place count be $n$ $n$
Let the $n^{th}$ $n^("th")$ term be $a_{n}$ $a_n$

Given $f (n = 1) = - 2$ $f(n=1)=-2$

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

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

From this the sequence is:

$n = 1 \to a_{1} = - 2 \leftarrow given value$ $n=1->a_1=-2 larr" given value"$

$n = 2 \to a_{2} = - 2 + 4 = 2$ $n=2->a_2=-2+4 = 2$
$n = 3 \to a_{3} = - 2 + 4 + 4 = 6$ $n=3->a_3=-2+4+4 = 6$
$n = 4 \to a_{4} = - 2 + 4 + 4 + 4 = 10$ $n=4->a_4=-2+4+4+4=10$

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

$a_{n} = - 2 + 4 (n - 1)$ $a_n =-2+4(n-1)$

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

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