How do you find nth term rule for #1,3,9,27,...#?

1 Answer
Azimet
Jan 31, 2017

The n-th term of this sequence appears to be #3^(n-1)#, #n >= 1#.

Explanation:

These are powers of #3# ordered from #3^0 = 1# to #3^a# (for an integer #a >=1#). However, the first convenient value for #n# is #1#, not #0# (imagine saying the 0th term of a sequence). Because of that, since the first term is actually #3^0#, we need to start from the first term (#n=1#) being

#3^(1-1)#. The next is #3^(2-1)#, #3^(3-1)# ... #3^(n-1)#.

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