Hi,
the moment you see the formula of nth number, it should tell you we are looking at a sequence ..
\(a_1 = \frac{1}{1}-\frac{1}{2}\)..
\(a_2= \frac{1}{2}-\frac{1}{3}\)..
\(a_3 = \frac{1}{3}-\frac{1}{4}\)..
.....
\(a_{99} = \frac{1}{99}-\frac{1}{100}\)..
\(a_{100} = \frac{1}{100}- \frac{1}{101}\)..
ADD all
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101} = 1-\frac{1}{101} = \frac{100}{101}\)
so when you add all terms, what is left is \(1-\frac{1}{101} = \frac{100}{101}\)..
E
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an= 1/n - 1/(n+1).

a1 = 1 - 1/2
a2 = 1/2 - 1/3
a3 = 1/3 - 1/4
.
.
.
a99 = 1/99 - 1/100
a100 = 1/100 - 1/101

Question: a1 + a2 + a3 + ...... + a99 + a100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......................... + 1/99 - 1/100 + 1/100 - 1/101 = 1 - 1/101 = 100/101

As you can see, except the first and the last terms the remaining terms get cancelled out.

Hope it helps.

No, don't use this formula here..
use it ONLY in an AP, where difference is common in consecutive numbers..
If you want to do approx, then too the terms should be atleast close to be called in AP
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This clearly look like a GP series

a1 = 1 - 1/2 = 1/2

a2 = 1/2 - 1/3 = 1/6

r = 1/3

S100 = 1/2 (( 1 - (1/3)^100)/(1 - 1/3))

= 3/4 * (1 - 1/3)^100)

------> is this correct? how to solve beyond this to reach the ans?
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Bunuel, chetan2u,

For me, the tricky part here is that the GMAT assumes that we know that (1- 1/2) + (1/2 - 1/4) + (1/4 - 1/8) ... (1/99-1/100)+(1/100-1/101) = 1
How can we know that? Is there any specific rule?

Many thanks in advance,

Jaime

Hi Vyshak,

Can you please explain in more detail? I still don't see it.

Many thanks in advance.

The sum formula that you have mentioned can be applied only when the sequence is in arithmetic progression. The given sequence is not in AP and hence cannot be applied.

The POE method (for non-genius like me):

a(n)= 1/n - 1/(n+1) => a(1)=1/2, a(2)=1/6, a(3)=1/12 ... a(100)=1/100*101

The largest is a(1) =1/2 and other terms become so small that they are not able to add together another 1/2 to overcome 1, but the sum is still +ve so crossout A,B,C.

Since we have to sum all this stuff and the last term a(100)=1/100*101 with prime number 101 in denominator, there is a huge chance that exactly 101 will stay in denominator. So the answer is (E).

p.s. This is not the best way to solve problems on GMAT, and you still have to learn the concept to be confident, but this is an example of how to make a smart guess and move on.