Why does a sum using the difference method need to factorise out common factors?

First I split #sum_1^n##1/(n(n+2))# into
#sum_1^n##1/(2n)-1/(2n+2)#
Then I input values to get
#(1/2-1/4 + 1/4 -1/6+...+1/(2n)-1/(2n+2))#

However this gives an incorrect answer, and the correct one is obtained only when the #1/2# is first factorised. Why is this the case?