Question 1121014
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I suspect the last fraction is typed incorrectly.  1/195 would make a much nicer problem.  I will solve the problem using 1/195 instead of 1/199.<br>
The series can be written as<br>
{{{1/(3*5)+1/(5*7)+1/(7*9)+1/(9*11)+1/(11*13)+1/(13*15)}}}<br>
Each term can be written in the form {{{1/((n-1)(n+1))}}}.<br>
Decomposing that fraction shows each term can be written in the form <br>
{{{(1/2)((1/(n-1)-1/(n+1)))}}}<br>
Then the sum is<br>
{{{(1/2)((1/3-1/5)+(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+(1/11-1/13)+(1/13-1/15))}}}
or
{{{(1/2)(1/3-1/15)}}}
which is
{{{(1/2)(4/15) = 2/15}}}<br>
If the last fraction is really 1/199, then treat that last fraction separately; the sum will be<br>
{{{(1/2)(1/3-1/13)+1/199}}}