Sequence And Series Gmat, Gre Question

The nth term of a sequence is given by an= 1/n - 1/(n+1) for every positive integer n. What is the sum of the first 100 terms?

(a) 1
(b) 0 
(c) 25
(d) 99/100
(e) 100/101





This may seem like an Arithmetic Progersion or even a Geometric Progression, but those are the eye-catching decoys. Add some terms to find out:
a1 = 1/2
a2 = 1/6
a3 = 1/12
a4 = 1/20
a5 = 1/30

Let's try a different approach:

term1 = 1/1 - 1/2 
term2 = 1/2 - 1/3  
...
term99 = 1/99 - 1/100
term100 = 1/100 - 1/101

So, the sum looks like this:

Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)  
Notice that most terms cancel out

Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101) 
=  1/1 - 1/101
= 101/101 - 1/101
= 100/101


E