Question

Find the first six terms of the sequence. a1 = 4, an = an-1 + 8

Answer 1

`\bf \begin{array}{llll} term&\stackrel{a_(n-1)+8}{value}\\ \cline{1-2} a_1&4\\ a_2&\stackrel{4+8}{12}\\ a_3&\stackrel{12+8}{20}\\ a_4&\stackrel{20+8}{28}\\ a_5&\stackrel{28+8}{36}\\ a_6&\stackrel{36+8}{44} \end{array}`

Answer 2

The first 6 terms are 4,12,20,28,36,44

Explanation:

So we have the recursive sequence

`a_n=a_(n-1)+8 \text{ with } a_1=4`.

If you try to dissect what this really means, it becomes easy.

Pretend
`a_n` is a term in your sequence.

Then
`a_(n-1)` is the term right before or something like
`a_(n+1)` means the term right after.

So it is telling us to find a term all we have to is add eight to the previous term.

So the second term
`a_2` is 4+8=12.

The third term is
`a_3` is 12+8=20.

The fourth term is
`a_4` is 20+8=28.

The fifth term is
`a_5` is 28+8=36

The sixth term is
`a_6` is 36+8=44.

Now sometimes it isn't that easy to see the pattern from the recursive definition of a relation. Sometimes the easiest way is to just plug in. Let's do a couple of rounds of that just to see what it looks like.

`a_n=a_(n-1)+8 \text{ with } a_1=4`.

`a_2=a_1+8=4+8=12`

`a_3=a_2+8=12+8=20`

`a_4=a_3+8=20+8=28`

`a_5=a_4+8=28+8=36`

`a_6=a_5+8=36+8=44`