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For every positive integer n, the nth term of sequence is given by an=

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inakihernandez
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For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  Updated on: 24 Apr 2016, 11:26
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For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
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Originally posted by inakihernandez on 24 Apr 2016, 10:38.
Last edited by Bunuel on 24 Apr 2016, 11:26, edited 1 time in total.
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chetan2u
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For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  24 Apr 2016, 10:47
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inakihernandez wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/n+1. What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


Hi,
the moment you see the formula of nth number, it should tell you we are looking at a sequence ..
\(a_1 = \frac{1}{1}-\frac{1}{2}\)..
\(a_2= \frac{1}{2}-\frac{1}{3}\)..
\(a_3 = \frac{1}{3}-\frac{1}{4}\)..
.....
\(a_{99} = \frac{1}{99}-\frac{1}{100}\)..
\(a_{100} = \frac{1}{100}- \frac{1}{101}\)..
ADD all
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101} = 1-\frac{1}{101} = \frac{100}{101}\)
so when you add all terms, what is left is \(1-\frac{1}{101} = \frac{100}{101}\)..
E
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BrentGMATPrepNow
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  28 Apr 2018, 07:11
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inakihernandez wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


term1 = 1/1 - 1/2
term2 = 1/2 - 1/3
term3 = 1/3 - 1/4
term4 = 1/4 - 1/5
.
.
.
term99 = 1/99 - 1/100
term100 = 1/100 - 1/101

So, the sum looks like this:

Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)

NOTICE THAT MOST TERMS CANCEL OUT

Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... + (1/99 - 1/100) + (1/100 - 1/101)
= 1/1 - 1/101
= 101/101 - 1/101
= 100/101
= E

Cheers,
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  24 Apr 2016, 11:21
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Hi chetan2u! thank you very much for your answer.

There is any way to calculate it using the formula (first term+last term)/2*number of terms?

because what I did as a first approach was a1= 1/2 ; a100= almost 0 => (1/2+0)/2*100= 25

I understood your solution but I was wondering if there is any way using the formula below, or was I using a wrong approach?
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  24 Apr 2016, 11:26
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inakihernandez wrote:
Hi chetan2u! thank you very much for your answer.

There is any way to calculate it using the formula (first term+last term)/2*number of terms?

because what I did as a first approach was a1= 1/2 ; a100= almost 0 => (1/2+0)/2*100= 25


No, don't use this formula here..
use it ONLY in an AP, where difference is common in consecutive numbers..
If you want to do approx, then too the terms should be atleast close to be called in AP
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  04 May 2016, 00:30
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inakihernandez wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


For similar questions, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/03 ... sequences/
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Re: gmat prep Q DS (series) [#permalink] New post  08 May 2016, 22:27
This clearly look like a GP series

a1 = 1 - 1/2 = 1/2

a2 = 1/2 - 1/3 = 1/6

r = 1/3

S100 = 1/2 (( 1 - (1/3)^100)/(1 - 1/3))

= 3/4 * (1 - 1/3)^100)

------> is this correct? how to solve beyond this to reach the ans?
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For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  08 May 2016, 22:54
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aniketm.87@gmail.com wrote:
For every positive integer n, the nth term of a sequence is given by an = 1/n - 1(n + 1). what is the sum of the 1st 100 terms in this sequence?

A) 0

B) 1/101

C) 99/100

D) 100/101

E) 1


hi
its not Gp..
At times we complicate the Q while simplifying it...

first write the numbers..
1st = \(1-\frac{1}{2}\)..
2nd = \(\frac{1}{2}-\frac{1}{3}\)
3rd = \(\frac{1}{3}-\frac{1}{4}\)..
99th = \(\frac{1}{99}-\frac{1}{100}\)..
1ooth = \(\frac{1}{100}-\frac{1}{101}\)...
add all ..
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101} = 1-\frac{1}{101} = \frac{100}{101}\)
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  14 May 2016, 14:43
Bunuel, chetan2u,

For me, the tricky part here is that the GMAT assumes that we know that (1- 1/2) + (1/2 - 1/4) + (1/4 - 1/8) ... (1/99-1/100)+(1/100-1/101) = 1
How can we know that? Is there any specific rule?

Many thanks in advance,

Jaime
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For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  14 May 2016, 20:09
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Jaikuz wrote:
Bunuel, chetan2u,

For me, the tricky part here is that the GMAT assumes that we know that (1- 1/2) + (1/2 - 1/4) + (1/4 - 1/8) ... (1/99-1/100)+(1/100-1/101) = 1
How can we know that? Is there any specific rule?

Many thanks in advance,

Jaime


Hi,

Except the first and the last term, all the other terms get cancelled out. So the only calculation required here is 1 - 1/101.
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  15 May 2016, 11:31
Hi Vyshak,

Can you please explain in more detail? I still don't see it.

Many thanks in advance.
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  15 May 2016, 12:13
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Jaikuz wrote:
Hi Vyshak,

Can you please explain in more detail? I still don't see it.

Many thanks in advance.


an= 1/n - 1/(n+1).

a1 = 1 - 1/2
a2 = 1/2 - 1/3
a3 = 1/3 - 1/4
.
.
.
a99 = 1/99 - 1/100
a100 = 1/100 - 1/101

Question: a1 + a2 + a3 + ...... + a99 + a100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......................... + 1/99 - 1/100 + 1/100 - 1/101 = 1 - 1/101 = 100/101

As you can see, except the first and the last terms the remaining terms get cancelled out.

Hope it helps.
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  30 Oct 2016, 10:25
Vyshak chetan2u : Why can't we use the Sum formula in this case? S100 = 100/2 * (2a + (100-1)d) ?

'Cuz if we do use, the answer comes to a value too far out from the answer choices, as per calculations.

Am i missing something?
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  30 Oct 2016, 10:41
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Vishvesh88 wrote:
Vyshak chetan2u : Why can't we use the Sum formula in this case? S100 = 100/2 * (2a + (100-1)d) ?

'Cuz if we do use, the answer comes to a value too far out from the answer choices, as per calculations.

Am i missing something?


The sum formula that you have mentioned can be applied only when the sequence is in arithmetic progression. The given sequence is not in AP and hence cannot be applied.
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  17 Dec 2017, 12:00
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Hello,

Adding information regarding why this is not an AP (arithmetic progression) nor a GP (geometric progression). Do the math for:
a1 = 1/2
a2 = 1/6
a3 = 1/12
a4 = 1/20
a5 = 1/30

If you try to put in any general term formula:
a2 = a1 + r.1 => r = -1/3
Trying to apply for a3: a3 = 1/12 <> 1/2 + 2.(-1/3)
The same goes with GP.

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For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  28 Apr 2018, 08:26
inakihernandez wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


The POE method (for non-genius like me):

a(n)= 1/n - 1/(n+1) => a(1)=1/2, a(2)=1/6, a(3)=1/12 ... a(100)=1/100*101

The largest is a(1) =1/2 and other terms become so small that they are not able to add together another 1/2 to overcome 1, but the sum is still +ve so crossout A,B,C.

Since we have to sum all this stuff and the last term a(100)=1/100*101 with prime number 101 in denominator, there is a huge chance that exactly 101 will stay in denominator. So the answer is (E).

p.s. This is not the best way to solve problems on GMAT, and you still have to learn the concept to be confident, but this is an example of how to make a smart guess and move on.
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For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  10 Jul 2019, 22:41
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I basically solved this via Hero8888 's approach in the test I did but I now understand what people mean by "they cancel each other out".

This needs to be better articulated.

DO NOT SOLVE for each term, instead write out the term before simplifying
i.e. a1 = 1 - 1/2
a2 = 1/2 -1/3
a100 = 1/100 - 1/101

Then adding the terms together means ADD THEM BEFORE SIMPLIFYING:
a1 + a2 + a3 ...+ a100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 ....+ 1/100 - 1/101

Now, everything from -1/2 to -1/100 (in term a99) will cancel out via addition and we are left with
1-1/101
=(101-1)/101 = 100/101
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink] New post  15 Jan 2021, 20:39
here is my alternative approach:

Quote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?



An = 1/n - 1/(n+1); can be solved to 1/n(n+1).

hence A100 = 1/100*101

The total numbers of terms are 100, hence the denominator has to be 101.
Answer E
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