Applying the formula, we get:
a1 = (1/1 - 1/3)
a2 = (1/2 - 1/4)
a3 = (1/3 - 1/5)
a4 = (1/4 - 1/6)
.
.
.
.
a17 = (1/17 - 1/19)
a18 = (1/18 - 1/20)
a19 = (1/19 - 1/21)
a20 = (1/20 - 1/22)

So, the SUM = (1/1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) . . . (1/17 - 1/19) + (1/18 - 1/20) + (1/19 - 1/21) + (1/20 - 1/22)

Notice that all of the SAME COLORED fractions cancel out.
For example, (-1/3) + 1/3 = 0

The only fractions the DON'T get canceled out are those at the beginning and end of the sequence.
So, SUM = 1/1 + 1/2 - 1/21 - 1/22
We can rewrite this as: SUM = (1/1 + 1/2) - (1/21 + 1/22)

Answer: B

Cheers,
Brent
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Here we have \(an= \frac{1}{n} - \frac{1}{(n+2)}\). The idea is to find a pattern in the sum

\(a1= \frac{1}{1} - \frac{1}{3}\)
\(a2= \frac{1}{2} - \frac{1}{4}\)
\(a3= \frac{1}{3} - \frac{1}{5}\)
\(a4= \frac{1}{4} - \frac{1}{6}\)

Now we have the first term from a3 can cancels second term from previous numbers

Now observing the last few terms...

\(a18= \frac{1}{18} - \frac{1}{20}\)
\(a19= \frac{1}{19} - \frac{1}{21}\)
\(a20= \frac{1}{20} - \frac{1}{22}\)

Now we have here the last two terms namely \(\frac{1}{21}\) and \(\frac{1}{22}\) are not cancelled.

\((1+\frac{1}{2}) - (\frac{1}{21}+\frac{1}{22})\)

Hence the sum is
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Now we have the first term from a3 can cancels second term from previous numbers
^^^ what do you mean by this exactly, like what exact numbers cancel?

does the sum of the first two cancel, or something else?

In the first set, you cancel both the 1/3, the 1/4, and the 1/5. You see the pattern now? This cancellation will continue until the last two terms (1/21 and 1/20 cannot be cancelled because n=20 is the limit).

In the first set you are left with 1/1 and 1/2. By the second set there's only 1/21 and 1/22. You add the first set and subtract the second. (1/1 + 1/2) - (1/21 + 1/22).

It took me 7 mins and 30 seconds to do this. How can someone do this within 2 mins or less?

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