A recursive rule for a geometric sequence is a1=4/9; an=3(an-1) What is an explicit rule for this sequence?
An =?
Are you familiar with that?
Yes
But when it comes to recursive terms, its a little confusing for me
No worries.. Step by step. \[\Huge a_1=\frac{ 4 }{ 9 }~~~,~~~ a_n=3*a_{n-1}\] That means two things: - the first term in this sequence is 4/9 - the common ration is 3, i.e to get any term you have to multiply the previous term by 3 to get the new term agree so far?
Yes, I catch on so far
What do I do next?
That is great! Now for the explicit rule for this sequence: you need to type out the first term \(\color{red}{a_1}\) in the sequence followed by the common ration \(\color{red}r\) raised to the power \(\color{red}{n-1}\) Why???? because when you are going to find out the first term, you would plug in n=1 into this \(a_n=a_1*r^{n-1}\), so you got the first term \(a_1\) only, so you can try yourself any other term
So what will that mean?
For the problem
\[\Huge a_n=\color{blue}{a_1}*\color{lime}r^{n-1}\] that means you plug in (4/9) instead of "the first term" \(a_1\) and 3 instead of the common ratio \(r\)
So how will that turn out?
It is your turn @Gamenerd123 now
Well, i mean like how do I do that, sorry.
Where is the first term? and where is the common ratio?
a1 is the first term and 3 is the common ratio, right?
a1=??
3?
if a1=3 and r=3, so 4/9=?????????????????
3
a_1=? r=?
Its not 3?
no it is not.. \[\Huge a_1=\frac{ 4 }{ 9 }~~~,~~~ a_n=3*a_{n-1}\] Do you remember that?
Yes
so?
Hmmm...
I really like who ask and stop the explanation when he sticks to something, even it looks so simple! So don't be embarrassed if there is anything not clear for you
My minds so slow for math I'm sorry
It takes a while for me to get it
No no don't say that ...I am not agree with you!
I have another idea...
Please, do tell..
Ok.. Can you type the formula of the explicit rule for this sequence?
Not exactly...
I'm very dumb for formulas
|dw:1482158022518:dw| got it?
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