Question 202544
I'll do the first one to get you going


# 1




Notice that each term is increasing exponentially. So this sequence might be a geometric sequence. To find out, let's simply divide the terms.



First divide the 2nd term 9 by the 1st term 3 to get  

{{{9/3=3}}} 

 
Now divide the 3rd term 27 by the 2nd term 9 to get  

{{{27/9=3}}} 

 
Now divide the 4th term 81 by the 3rd term 27 to get  

{{{81/27=3}}} 

 
Now divide the 5th term 243 by the 4th term 81 to get  

{{{243/81=3}}} 

 

So if we pick ANY term and divide it by the previous term, we'll always get 3. 


This is the common ratio between the terms. So this means that {{{r=3}}}.




Now since we've started at 3, this means that {{{a=3}}}




Since the general geometric sequence is {{{a[n]=ar^n}}}, this means the sequence is


{{{a[n]=3*3^n}}}  where {{{n}}} starts at {{{n=0}}}



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Answer:


So the sequence is {{{a[n]=3*3^n}}} where {{{n}}} starts at {{{n=0}}}


Note: if you want to start at {{{n=1}}}, then you need to replace "n" with "n-1" to get the new sequence {{{a[n]=3*3^(n-1)}}}