Question 1199206
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Simplify: {{{(1/4) + (1/4^2) + ((4^2)/(4^3)) + ((4^2)/(4^4)) + ((4^4)/(4^5)) + ((4^4)/(4^6))}}} + ... + {{{((4^102)/(4^103)) + ((4^102)/(4^104))}}}
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<pre>
I see two sequences alternate.


After separating, one sequence is   {{{4^0/4^1}}} + {{{4^2/4^3}}} + {{{4^4/4^5}}} + . . . + {{{4^102/4^103}}}         (1)


              Another sequence is   {{{4^0/4^2}}} + {{{4^2/4^4}}} + {{{4^6/4^8}}} + . . . + {{{4^102/4^104}}}         (2)


Sequence (1), after reducing, is the sum of {{{102/2+1}}} = 52 equal addends, each of which is  {{{1/4}}},
so this sum is  {{{52/4}}}.


Sequence (2), after reducing, is the sum of {{{102/2+1}}} = 52 equal addends, each of which is  {{{1/16}}},
so this sum is  {{{52/16}}}.


Therefore, the total sum is  {{{52/4}}} + {{{52/16}}} = {{{(52*4)/16}}} + {{{52/16}}} = {{{260/16}}} = {{{130/8}}} = 16{{{2/8}}} = 16{{{1/4}}}.    <U>ANSWER</U>
</pre>


Solved.