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What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9

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Nevernevergiveup
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What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   Updated on: 18 Feb 2019, 05:33
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What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?


A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)
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E

Originally posted by Nevernevergiveup on 16 Apr 2016, 10:57.
Last edited by Bunuel on 18 Feb 2019, 05:33, edited 1 time in total.
Updated.
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   16 Apr 2016, 11:17
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Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


the answer is E.
We have 1/n(n+1) = 1/n - 1/(n+1)
So, \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} = 1- \frac{1}{10} = \frac{9}{10}\). :)
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   25 Jul 2018, 22:07
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Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)
.

The given expression is of the form \(\frac{1}{n(n+1)}\) = \(\frac{1}{n}- \frac{1}{(n+1)}\).

So, the expression can be re-written as
\(1\\
-\frac{1}{(2)}+\frac{1}{(2)}\\
-\frac{1}{(3)}+\frac{1}{(3)}\\
-\frac{1}{(4)}+\frac{1}{(4)}\\
-\frac{1}{(5)}+\frac{1}{(5)}\\
-\frac{1}{(6)}+\frac{1}{(6)}\\
-\frac{1}{(7)}+\frac{1}{(7)}\\
-\frac{1}{(8)}+\frac{1}{(8)}\\
-\frac{1}{(9)}+\frac{1}{(9)}\\
-\frac{1}{(10)}\)
= \(1- \frac{1}{10}\) = \(\frac{9}{10}\)

Ans - E.
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   16 Apr 2016, 16:38
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to find the sum of n terms in this series,
add n-1 to both the numerator and denominator of term 1
term 1=1/2
for summing 9 terms, n-1=8
(1+8)/(2+8)=9/10
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   14 Nov 2016, 09:08
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Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)



I had a slightly different approach...
1/2 + 1/6 + 1/12 + 1/20 + 1/30
we can find LCM = 120
(60+20+10+6+4)/120 = 100/120 = 10/12 = 5/6

1/6 = 0.16
5/6 = 0.83+++
so just only first 5 fractions add up to 0.83...

A, B, C are out.
now, between D and E
D = 46/55 - this is equal to 0.83 and smth
we need an answer that is greater than 0.83

answer must be E.
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   15 Nov 2016, 12:51
anhnguyen178 wrote:
Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


the answer is E.
We have 1/n(n+1) = 1/n - 1/(n+1)
So, \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} = 1- \frac{1}{10} = \frac{9}{10}\). :)



Could you please elaborate more on that? I can't understand wha you have applied
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   15 Nov 2016, 15:21
Are there any formulas/approaches to this question instead of brute force math? I have an issue with summation problems when it comes to descending fraction-values..
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   17 Nov 2016, 15:15
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This question is more about seeing a pattern than actually doing math.

For example, you should notice once you start adding the following:

(1/2)+(1/6)=(4/6)=(2/3) --> The sum of the first two is equal to the (n-1)/n of the 2nd value given in the prompt

If you're not convinced, let's try adding the next fraction: (4/6)+(1/12)=(3/4) ...(n-1)/n of 3rd value

Thus, (1/2)+....+(1/90) = 9/10

E
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   23 Nov 2016, 08:39
mvictor wrote:
Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)



I had a slightly different approach...
1/2 + 1/6 + 1/12 + 1/20 + 1/30
we can find LCM = 120
(60+20+10+6+4)/120 = 100/120 = 10/12 = 5/6

1/6 = 0.16
5/6 = 0.83+++
so just only first 5 fractions add up to 0.83...

A, B, C are out.
now, between D and E
D = 46/55 - this is equal to 0.83 and smth
we need an answer that is greater than 0.83

answer must be E.



solved it again...almost same method used...
i transformed everything into 1/900
1/2 = 450/900
1/6 = 150/900
1/12 = 75/900
1/20 = 45/900
1/30 = 30/900
1/42 = slightly more than 21/900
1/56 = slightly more than 16/900
1/72 = slightly more than 10/900
1/90 = 10/900

450+150 = 600
75+45 = 120
21+16+10+10+30 = 87
600+120+87 = 807/900
so it must definitely be >8/9, which is 0.88888....A/B/C are out right away
46/55 = 92/110 -> or aprox 9/11. 1/11 = 0.09 -> * 9 ~ 0.82, which is less than needed.

E must be the answer.
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   28 Nov 2016, 15:41
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Here we just have to split each term
1/2=> 1-1/2
1/6=1/2-1/6
1/12=1/4-1/3
.
.
.
.
1/90=1/9-1/10

As we can see when we add them => The second term of each term (except the last) camels out the first term in each term (except the first)

Hence we are left with => 1-1/10 => 9/10

Hence E
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What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   18 Aug 2019, 09:40
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Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?


A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


Asked: What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

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

\(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} =\)

\(1-\frac{1}{(2)}+\frac{1}{(2)}\\
-\frac{1}{(3)}+\frac{1}{(3)}\\
-\frac{1}{(4)}+\frac{1}{(4)}\\
-\frac{1}{(5)}+\frac{1}{(5)}\\
-\frac{1}{(6)}+\frac{1}{(6)}\\
-\frac{1}{(7)}+\frac{1}{(7)}\\
-\frac{1}{(8)}+\frac{1}{(8)}\\
-\frac{1}{(9)}+\frac{1}{(9)}\\
-\frac{1}{(10)}= 1 - \frac{1}{10} = \frac{9}{10}\)

IMO E
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   18 Aug 2019, 10:15
The answer has to be more than (1/2 + 1/6 + 1/12) i.e more than (0.5 + 0.16 + 0.08) i.e. more than 0.74
Only option remains is E

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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   18 Aug 2019, 12:42
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Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?


A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


I basically took the nice fractions:
1/2+1/6+1/12+1/20 which gives me .5+.17+.08+.05 (You can think of them as percentages as well). This itself gives you .8. If you approximate the other fractions you realise you're going towards a total of 0.9 Hence E.
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   08 May 2021, 03:49
anhnguyen178 wrote:
Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


the answer is E.
We have 1/n(n+1) = 1/n - 1/(n+1)
So, \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} = 1- \frac{1}{10} = \frac{9}{10}\). :)


VeritasKarishma can you pls explain how to arrive at \(1- \frac{1}{10} \)
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   08 May 2021, 03:55
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dave13 wrote:
anhnguyen178 wrote:
Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


the answer is E.
We have 1/n(n+1) = 1/n - 1/(n+1)
So, \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} = 1- \frac{1}{10} = \frac{9}{10}\). :)


VeritasKarishma can you pls explain how to arrive at \(1- \frac{1}{10} \)


Hi dave13
The given expression is of the form \(\frac{1}{n(n+1)}\) = \(\frac{1}{n}- \frac{1}{(n+1)}\).

So, the expression can be re-written as
\(1\)\(-\frac{1}{(2)}+\frac{1}{(2)}\)\(-\frac{1}{(3)}+\frac{1}{(3)}\)\(-\frac{1}{(4)}+\frac{1}{(4)}\)\(-\frac{1}{(5)}+\frac{1}{(5)}\)\(-\frac{1}{(6)}+\frac{1}{(6)}\)\(-\frac{1}{(7)}+\frac{1}{(7)}\)\(-\frac{1}{(8)}+\frac{1}{(8)}\)\(-\frac{1}{(9)}+\frac{1}{(9)}\)\(-\frac{1}{(10)}\)

So all the colored terms cancel each other respectively and what we are left with is

= \(1- \frac{1}{10}\) = \(\frac{9}{10}\)

I hope this help! :)
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   08 May 2021, 04:25
many thanks GMATinsight but i still have question :lol: how did you figure out that it is in the form of

\(\frac{1}{n}- \frac{1}{(n+1)}\). :? also why are you subtracting one fraction from another ....whereas all fraction are added in the question
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   08 May 2021, 04:52
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dave13 wrote:
many thanks GMATinsight but i still have question :lol: how did you figure out that it is in the form of

\(\frac{1}{n}- \frac{1}{(n+1)}\). :? also why are you subtracting one fraction from another ....whereas all fraction are added in the question


EXPERIENCE and EXPOSURE :D

Next time you too will know... ;) :cool:
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   10 May 2021, 00:32
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dave13 wrote:
anhnguyen178 wrote:
Nevernevergiveup wrote:
What is \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)}\)?

A. \(\frac{2}{5}\)

B. \(\frac{3}{5}\)

C. \(\frac{7}{10}\)

D. \(\frac{46}{55}\)

E. \(\frac{9}{10}\)


the answer is E.
We have 1/n(n+1) = 1/n - 1/(n+1)
So, \(\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} = 1- \frac{1}{10} = \frac{9}{10}\). :)


VeritasKarishma can you pls explain how to arrive at \(1- \frac{1}{10} \)


Don't sweat too much about it. It is not a GMAT type question. It is a standard question type for some Indian tests. Though no harm in getting the exposure.
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink] New post   23 Aug 2023, 10:49
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Re: What is 1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9 [#permalink]
23 Aug 2023, 10:49
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