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# sequence of numbers a1, a2, a3, . . . , an

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sequence of numbers a1, a2, a3, . . . , an [#permalink]  04 Jan 2016, 06:20
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55% (01:15) correct 45% (02:09) wrong based on 20 sessions
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. $$\frac{1}{20} - \frac{1}{22}$$

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

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Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1825 [0], given: 397

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]  04 Jan 2016, 08:40
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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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]  29 Mar 2016, 09:34
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?
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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]  13 Jul 2016, 13:31
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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).
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Re: sequence of numbers a1, a2, a3, . . . , an [#permalink]  09 Feb 2019, 08:17
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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

Re: sequence of numbers a1, a2, a3, . . . , an   [#permalink] 09 Feb 2019, 08:17
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# sequence of numbers a1, a2, a3, . . . , an

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