Respuesta :
[tex]\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}[/tex]
Answer:
The first 6 terms are 4,12,20,28,36,44
Step-by-step explanation:
So we have the recursive sequence
[tex]a_n=a_{n-1}+8 \text{ with } a_1=4[/tex].
If you try to dissect what this really means, it becomes easy.
Pretend [tex]a_n[/tex] is a term in your sequence.
Then [tex]a_{n-1}[/tex] is the term right before or something like [tex]a_{n+1}[/tex] 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 [tex]a_2[/tex] is 4+8=12.
The third term is [tex]a_3[/tex] is 12+8=20.
The fourth term is [tex]a_4[/tex] is 20+8=28.
The fifth term is [tex]a_5[/tex] is 28+8=36
The sixth term is [tex]a_6[/tex] 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.
[tex]a_n=a_{n-1}+8 \text{ with } a_1=4[/tex].
[tex]a_2=a_1+8=4+8=12[/tex]
[tex]a_3=a_2+8=12+8=20[/tex]
[tex]a_4=a_3+8=20+8=28[/tex]
[tex]a_5=a_4+8=28+8=36[/tex]
[tex]a_6=a_5+8=36+8=44[/tex]
