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The sequence x1, x2, x3,..., is such that Xn = 1/n - 1/(n+1)

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Intern
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Joined: 04 Sep 2009
Posts: 43
Followers: 1

Kudos [?]: 8 [0], given: 9

The sequence x1, x2, x3,..., is such that Xn = 1/n - 1/(n+1) [#permalink] New post 22 Sep 2012, 09:19
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (02:23) correct 34% (01:40) wrong based on 73 sessions
The sequence x1, x2, x3,..., is such that Xn = 1/n - (1/(n+1)). What is the sum of the first 100 terms of the sequence?

A. 201/100
B. 99/100
C. 100/101
D. 1/10000
E. 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Sep 2012, 09:31, edited 1 time in total.
Renamed the topic and edited the question.
Intern
Intern
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Joined: 02 Nov 2009
Posts: 44
Location: India
Concentration: General Management, Technology
GMAT Date: 04-21-2013
GPA: 4
WE: Information Technology (Internet and New Media)
Followers: 3

Kudos [?]: 35 [0], given: 8

Re: The sequence x1, x2, x3,..., is such that Xn = 1/n - 1/(n+1) [#permalink] New post 22 Sep 2012, 10:18
Given Xn = 1/n - (1/(n+1))
--> X1=1-(1/2)
X2=(1/2)-(1/3)
X3=(1/3)-(1/4)

x1+x2+x3 = 1-(1/4) as other terms gets cancelled out...


following the same approach x1+x2...+x100 =1-(1/101) =100/101

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Concentration: Technology, Finance
GMAT 1: 580 Q48 V21
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Kudos [?]: 23 [0], given: 25

Re: The sequence x1, x2, x3,..., is such that Xn = 1/n - 1/(n+1) [#permalink] New post 20 Nov 2013, 10:50
Easy task and accomplish

X1=1-1/2
x2=1/2-1/3
x3=1/3-1/4
.....
x99=1/99-1/100
x100=1/100-1/101

sum=X1+X2+X3+....X100=1-1/2+1/2-1/3+.......1/99-1/100+1/100-1/101=1-1/101=100/101

c is the answer
.
.
Re: The sequence x1, x2, x3,..., is such that Xn = 1/n - 1/(n+1)   [#permalink] 20 Nov 2013, 10:50
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The sequence x1, x2, x3,..., is such that Xn = 1/n - 1/(n+1)

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