What is the nth term of 1,2,4,8,16,32?

1 Answer
MeneerNask
Mar 21, 2016

It's apparent that the first term is #1# and the ratio is #2#

Explanation:

In 'the language':
#u_0=1,r=2#

Then:
#u_n=u_0xxr^n#

But:
This is not the #n#th term, as the first one is called #u_0#
So you need #u_(n-1)=u_0xxr^(n-1)#

Example:
The tenth term would be nine steps away from the first (called #0#), which would translate to #u_9=1xx2^9=512#

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