Question 742212
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I need to find the sum of the following infinite series: ∑[n-0,∞,1/4^n]

(The n-0 is below the ∑, and the ∞ is above. The 1/4^n is one the right side of the ∑)
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This problem is to find the sum of the infinite geometric progression 

having the first term  1 = {{{(1/4)^0}}}  and the common ratio  {{{1/4}}}.


The general formula is  S = {{{a/(1-r)}}},  where 'a'  is the first term of a GP,
and  'r'  is the common ratio.


For this problem, the sum is  S = {{{1/(1-1/4)}}} = {{{1/((3/4))}}} = {{{4/3}}}.


<U>ANSWER</U>.  S = {{{4/3}}}.
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Solved.