Question 1117401
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<pre>
{{{2/(x*(x+2))}}} = {{{1/x}}} - {{{1/(x+2)}}}.


So, 


  {{{2/(1*3)}}}  + {{{2/(3*5)}}} + {{{2/(5*7)}}}  + ...   + {{{2/(99*101)}}} = 


= {{{(1/1 - 1/3)}}} + {{{1/3 - 1/5)}}} + {{{(1/5-1/7)}}} + . . . + {{{(1/99-1/101)}}} = 


= all intermediate terms cancel each other, which leaves = {{{1 - 1/101}}} = {{{100/101}}}.
</pre>

Solved.


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These tricks are very well known.  &nbsp;See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/NumericFractions/Calculations-with-fractions.lesson>Calculations with fractions</A>

in this site.