Question 70074
<pre><font size = 5><b>the first example in a lecture of series is a^n=1/2^n
can you write the recursion formula for this series? 
please help with this, thanks

a<sub>n+1</sub> = {{{ 1/(2^(n+1)) }}}={{{ 1/(2^n*2^1)=(1/(2^n))(1/(2^1))}}}=(a<sub>n</sub>){{{(1/2)}}} = 

          a<sub>n</sub>
a<sub>n+1</sub> =   ————         
          2                  

Edwin</pre>