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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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Bunuel
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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink] New post   20 Feb 2017, 02:01
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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
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C
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For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink] New post   20 Feb 2017, 02:20
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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] New post   20 Feb 2017, 02:27
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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] New post   21 Jan 2018, 14:24
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] New post   30 Nov 2018, 07:43
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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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink] New post   26 Jun 2019, 08:51
Bunuel

Are you sure this is a Veritas Prep question?

I am fairly certain that this question is actually from the GMAT Prep EP2 or a pretty blatant copy of an OG question.

https://gmatclub.com/forum/for-every-po ... 17215.html

Best regards,
Chris
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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink] New post   26 Jun 2019, 09:12
100/101
--->

a1= 1/1- 1/2
a2 = 1/2 -1/3
a3 = 1/3 -1/4
a4 = 1/4 -1/5
.
.
.
a100=1/100-1/101

the second term of the previous term and the first term of the present term get cancelled out (e.g 1/2 from a1 and 1/2 from a3 get cancelled out)
... leaving only 1/1 and 100/101

Thus, 1/1-1/101 = 101/101-1/101 = 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] New post   01 Jul 2019, 17:14
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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


Let’s calculate the first few terms:

a1 = 1/1 - 1/2

a2 = 1/2 - 1/3

a3 = 1/3 - 1/4

a4 = 1/4 - 1/5

Summing the first 4 terms of the sequence, we have 1 - 1/5 because all the middle numbers cancel. Thus, we see that only the first and last term are left. So the sum of the first 100 terms would be 1 - 1/101 = 100/101.

Answer: C
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Re: For the sequence a1, a2, a3 ... an, an is defined by a_n=1/n−1/(n+1) [#permalink] New post   05 Aug 2021, 10:59
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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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