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A sequence is such that, an=1/n-1/(n+1). What is the sum of [#permalink ]
22 Mar 2006, 17:27

A sequence is such that, an=1/n-1/(n+1). What is the sum of the
first 100 terms?

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joemama142000 wrote:

A sequence is such that, an=1/n-1/(n+1). What is the sum of the first 100 terms?

hmm... interesting..

a1=1/1-1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

........

.........

.........

a100 = 1/100 - 1/101

so, sum = 1 - 1/101 = 100/101

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good solution professor..
I like the simplicity.. For some reason I started using Arithematic Progression formulaes... And it did not work with the formulaes..
any ideas why?

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Professor wrote:

joemama142000 wrote:

A sequence is such that, an=1/n-1/(n+1). What is the sum of the first 100 terms?

hmm... interesting..

a1=1/1-1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

........

.........

.........

a100 = 1/100 - 1/101

so, sum = 1 - 1/101 = 100/101

prof, can you please explain the part

1- 1/101= 100/101

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Angela780 wrote:

Professor wrote:

joemama142000 wrote:

A sequence is such that, an=1/n-1/(n+1). What is the sum of the first 100 terms?

hmm... interesting..

a1=1/1-1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

........

.........

.........

a100 = 1/100 - 1/101

so, sum = 1 - 1/101 = 100/101

prof, can you please explain the part

1- 1/101= 100/101

a1 + a2 + a3 + a4 +....

all the terms cancel out from a2 to a99 because of the alternative negative and postive signs

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Ok, thank you for the explanation.

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Angela780 wrote:

Professor wrote:

joemama142000 wrote:

A sequence is such that, an=1/n-1/(n+1). What is the sum of the first 100 terms?

hmm... interesting..

a1=1/1-1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

........

.........

.........

a100 = 1/100 - 1/101

so, sum = 1 - 1/101 = 100/101

prof, can you please explain the part

1- 1/101= 100/101

= 1- 1/101 = (1) - (1)/(101) = (101-1)/(101) = 100/101

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oa is 100/101 great work prof

Manager

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Professor wrote:

Angela780 wrote:

Professor wrote:

joemama142000 wrote:

A sequence is such that, an=1/n-1/(n+1). What is the sum of the first 100 terms?

hmm... interesting..

a1=1/1-1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

........

.........

.........

a100 = 1/100 - 1/101

so, sum = 1 - 1/101 = 100/101

prof, can you please explain the part

1- 1/101= 100/101

= 1- 1/101 = (1) - (1)/(101) = (101-1)/(101) = 100/101

I'm not sure why you take 1-1/101... could you explain why? I read that it cancels out but I still don't see how it cancels?

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Quote:

1 - 1/101

I also don't exactly understand how you get to the above. Please explain in more detail. Thanks.

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Charlie45 wrote:

Quote:

1 - 1/101

I also don't exactly understand how you get to the above. Please explain in more detail. Thanks.

a1 = 1/1 - 1/2

a2 = 1/2 - 1/3

a3 = 1/3 - 1/4

Just taking sum of a1 + a2 + a3, we have

1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 = 1/1 - 1/4.

You can see that the middle terms all cancel out leaving you with the first term and last term.

Just one more term, in case you're not convinced...

a4 = 1/4 - 1/5

so a1 + a2 + a3 + a4 + a5 = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 = 1/1 - 1/5

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Thanks ywilfred, very clever solution.

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ur a genious!

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Professor wrote:

joemama142000 wrote:

A sequence is such that, an=1/n-1/(n+1). What is the sum of the first 100 terms?

hmm... interesting..

a1=1/1-1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

........

.........

.........

a100 = 1/100 - 1/101

so, sum = 1 - 1/101 = 100/101

all i can say that you are super genius........