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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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sequence of numbers a1, a2, a3, . . . , an [#permalink]
Expert's post 00:00

Question Stats: 58% (01:37) correct 41% (02:26) wrong based on 31 sessions
The sequence of numbers $$a_1$$, $$a_2$$, $$a_3$$, . . . , $$a_n$$, . . . is defined by $$a_n$$$$= \frac{1}{n} - \frac{1}{(n+2)}$$ for each integer $$n ≥ 1$$. What is the sum of the first 20 terms of this sequence?

A. $$(1+\frac{1}{2}) - \frac{1}{20}$$

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

C. $$1 - (\frac{1}{20}+\frac{1}{22})$$

D. $$1 - \frac{1}{22}$$

E. $$\frac{1}{20} - \frac{1}{22}$$

Practice Questions
Question: 12
Page: 158
Difficulty: hard
[Reveal] Spoiler: OA

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Sandy
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 135

Kudos [?]: 2083 , given: 397

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
Expert's post
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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Sandy
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Intern Joined: 28 Mar 2016
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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
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? Intern Joined: 12 Jul 2016
Posts: 8
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Kudos [?]: 4  , given: 11

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
1
KUDOS
sagnik242 wrote:
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). GRE Instructor Joined: 10 Apr 2015
Posts: 2386
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Kudos [?]: 2353  , given: 33

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]
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Expert's post
sandy wrote:
The sequence of numbers a1, a2, a3, . . . , an, . . . is defined by $$an= \frac{1}{n} - \frac{1}{(n+2)}$$ for each integer n ≥ 1. What is the sum of the first 20 terms of this sequence?

A. $$(1+\frac{1}{2}) - \frac{1}{20}$$
B. $$(1+\frac{1}{2}) - (\frac{1}{21}+\frac{1}{22})$$
C. $$1 - (\frac{1}{20}+\frac{1}{22})$$
D. $$1 - \frac{1}{22}$$
E. $$\frac{1}{20} - \frac{1}{22}$$

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)

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails Re: sequence of numbers a1, a2, a3, . . . , an   [#permalink] 09 Feb 2019, 08:17
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