find the sum of n terms: 1/1.2.3 + 1/2.3.4 + 1/3.4.5 ....

The general term is 1/(n-1)n(n+1).

After playing with this a little, it looks like:
1/(n-1)n(n+1) = 
(1/2) [1/(n-1)n - 1/n(n+1)]

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nth term=1/(n)(n+1)(n+2)

equate n+1=0

implies,n=-1

substitute in n+2

-1+2   =1

similarly

equate n+2=0

implies n=-2

substitute the value of n in n+1

=-2+1     =-1

man the answer doesnt seem right, if u enter n=1, the sum to n terms comes to be negative..

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