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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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Math Expert
Joined: 02 Sep 2009
Posts: 43894
For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink]

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20 Feb 2017, 01:01
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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
[Reveal] Spoiler: OA

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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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20 Feb 2017, 01:20
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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.
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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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20 Feb 2017, 01:27
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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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Joined: 21 Jan 2018
Posts: 1
Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink]

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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))...
Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1)   [#permalink] 21 Jan 2018, 13:24
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