2012/011) The sum of 1+1/3+1/6+1/10+1/15......2005 terms is

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

we can check on that from 1,3,6,10 the common difference is 2,3,4,5...

1st tern = 2/1 - 2/2
2nd term = 2/2- 2/3
2005th term = 2/2005 - 2/2006

sum up to 2005 terms is 2- 2/2006 = (2006*2-2)/2006 = 2005/1003

note:
sum up to n terms = 2- 2/(n+1) = 2(1- 1/(n+1)) = 2n/(n+1)