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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)

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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)  [#permalink]

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New post 20 Feb 2017, 01:01
1
1
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (01:27) correct 36% (02:07) wrong based on 118 sessions

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For the sequence a1, a2, a3 ... an, an is defined by \(a_n=\frac{1}{n}−\frac{1}{(n+1)}\) for each integer n≥1. What is the sum of the first 100 terms of this sequence?

A. 100/99

B. 101/100

C. 100/101

D. 99/100

E. 999/1010

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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)  [#permalink]

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New post 20 Feb 2017, 01:20
3
Bunuel wrote:
For the sequence a1, a2, a3 ... an, an is defined by \(a_n=\frac{1}{n}−\frac{1}{(n+1)}\) for each integer n≥1. What is the sum of the first 100 terms of this sequence?

A. 100/99

B. 101/100

C. 100/101

D. 99/100

E. 999/1010


\(a_n=\frac{1}{n}−\frac{1}{(n+1)}\)

\(a_1 + a_2 + a_3 + a_4 + ......+ a_99 + a_100=\frac{1}{1}−\frac{1}{2}+\frac{1}{2}−\frac{1}{3}+\frac{1}{3}−\frac{1}{4}+\frac{1}{4}−\frac{1}{5}+.....+ \frac{1}{99}−\frac{1}{100}+\frac{1}{100}−\frac{1}{101}\)

Everything in between 1/1 and -(1/101) gets cancelled out.

\(a_1 + a_2 + a_3 + a_4 + ......+ a_99 + a_100= \frac{1}{1}−\frac{1}{101} = \frac{100}{101}\)

Hence Option C is correct.
Hit Kudos if you liked it 8-)
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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)  [#permalink]

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New post 20 Feb 2017, 01:27
1
C
Coz
a1= 1- 1/2
a2 = 1/2 -1/3

a100= 1/100 - 1/101
Adding all d above we get

1-1/101 = 100/101

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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)  [#permalink]

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New post 21 Jan 2018, 13:24
Just curious why the each term would not be included in parenthesis? Wouldn't this change the order of operations and the answer?

(A1)+(a2)...
1/2+((1/2)-(1/3))+((1/3)-(1/4))...
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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)  [#permalink]

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New post 30 Nov 2018, 06:43
I didn't pick up on the pattern while calculating a1, a2, a3; however, I did a couple of sums and picked up on the pattern then:
a1=1/2
a1+a2=2/3
a1+a2+a3=3/4

so a1+a2+....+a100=100/100+1=100/101
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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) &nbs [#permalink] 30 Nov 2018, 06:43
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