Question 1035284
This is what you call a telescoping series.

{{{sum(1/(r(r+2)), r=1, n) = sum((1/2)(1/r - 1/(r+2)), r=1, n)}}}

=  {{{1/2}}}({{{1-1/3 +1/2 - 1/4 + 1/3 - 1/5}}}+...+{{{1/(n-2)-1/n + 1/(n-1) - 1/(n+1) + 1/n - 1/(n+2)}}})

={{{(1/2)(1 + 1/2 -1/(n+1) - 1/(n+2))}}}

= {{{3/4 - (2n+3)/(2n^2+6n+4)}}}