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?
First I split
Then I input values to get
However this gives an incorrect answer, and the correct one is obtained only when the
1 Answer
Mar 20, 2018
You have an error.
