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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/

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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/  [#permalink]

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Updated on: 13 Mar 2019, 04:32
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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. 2/5
B. 3/5
C. 7/10
D. 46/55
E. 9/10

Originally posted by goodyear2013 on 18 Mar 2014, 14:59.
Last edited by Bunuel on 13 Mar 2019, 04:32, edited 1 time in total.
Renamed the topic and edited the question.
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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/  [#permalink]

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17 Jun 2015, 23:36
25
1
9
Method-1

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

Can be Rewritten as

$$[\frac{1}{1}-\frac{1}{2}] + [\frac{1}{2}-\frac{1}{3}] + [\frac{1}{3}-\frac{1}{4}] + [\frac{1}{4}-\frac{1}{5}] + [\frac{1}{5}-\frac{1}{6}] + [\frac{1}{6}-\frac{1}{7}] + [\frac{1}{7}-\frac{1}{8}] + [\frac{1}{8}-\frac{1}{9}] + [\frac{1}{9}-\frac{1}{10}]$$

Every terms cancels out except

$$[\frac{1}{1}-\frac{1}{10}] = \frac{9}{10}$$
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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/  [#permalink]

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19 Mar 2014, 10:51
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Tn = nth term of the series = 1/(n)(n+1) = (n+1-n)/(n)(n+1) = 1/n - 1/(n+1)

If you add all the terms of this series (to n terms), you get a sum of 1 - 1/(n+1) as all the other terms cancel out.
So sum of 9 terms = 1 - 1/10 = 9/10.
Option (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/  [#permalink]

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17 Jun 2015, 23:42
3
Method-2

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

Can be Rewritten as

$$0.5 + 0.166 + 0.083 + 0.05 + 0.03 + 0.023 +...$$ and so on (Other values are too small)

SUM = Approximately $$0.85$$ and a little more

Check Options:

A) 2/5 = 0.4
B) 3/5 = 0.6
C) 7/10 = 0.7
D) 46/55 = 0.83
E) 9/10 = 0.9 CORRECT

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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/  [#permalink]

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18 Mar 2014, 22:53
2
1
goodyear2013 wrote:
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) + 1/(9)*(10)?

2/5
3/5
7/10
46/55
9/10

Hi,
I want to know what is the best way to solve this question in 2min, please.

I would not worry much about the values of fractions after 1/7*8 as they will have minimal effect on the final answer. It is also important to recognize pattern after first few divisions
48027
So, 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) + /1(5)*(6) + 1/(6)*(7) ----------> 0.5 + 0.16 + 0.08 + 0.05 + 0.03 + ....... Looking at this pattern, it is safe to add 0.2. and 0.1 in the addition -----> 0.85. We can conclude that answer must be grater than 0.85. Answer E is slightly greater than this. It took me 2.32 min to solve this problem, though.
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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/  [#permalink]

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29 Jun 2015, 09:43
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Just want to post a general strategy to solve this kind of question. As this is already solved, I will try on another similar kind. Let's try to find out the value of the below expression:

$$\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}$$

The key to solve these type of questions is to rewrite each term as difference of two fractions and then cancel out maximum possible terms.

How can we rewrite each term as difference of two fractions??

1st term : $$\frac{1}{1*3} = (\frac{1}{2})\frac{2}{1*3} = (\frac{1}{2})\frac{3-1}{1*3} = (\frac{1}{2})(\frac{3}{1*3} - \frac{1}{1*3}) = (\frac{1}{2})(\frac{1}{1} - \frac{1}{3})$$

2nd term : $$\frac{1}{2*4} = (\frac{1}{2})\frac{2}{2*4} = (\frac{1}{2})\frac{4-2}{2*4} = (\frac{1}{2})(\frac{4}{2*4} - \frac{2}{2*4}) = (\frac{1}{2})(\frac{1}{2} - \frac{1}{4})$$

Similarly, 3rd term: $$\frac{1}{3*5} = (\frac{1}{2})(\frac{1}{3} - \frac{1}{5})$$

(n-1)th term: $$\frac{1}{(n-1)*(n+1)} = (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)})$$

nth term: $$\frac{1}{n*(n+2)} = (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})$$

Now using the above terms in the expression:

$$\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}$$

= $$(\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) +(\frac{1}{2})(\frac{1}{4} - \frac{1}{6}) + ........ + (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)}) + (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})$$

= $$(\frac{1}{2}) (\frac{1}{1}$$ $$- \frac{1}{3}$$ $$+ \frac{1}{2}$$ $$- \frac{1}{4}$$ $$+ \frac{1}{3}$$ $$- \frac{1}{5}$$ + $$\frac{1}{4}$$ $$- \frac{1}{6} + ....... + \frac{1}{(n-2)}$$ $$- \frac{1}{n}$$ + $$\frac{1}{(n-1)} - \frac{1}{(n+1)}$$ $$+ \frac{1}{n}$$ $$- \frac{1}{(n+2)})$$

In the above expression, you can notice that the same coloured terms can be cancelled out.

Now after cancelling we can simplify the complete expression to

$$(\frac{1}{2}) (\frac{1}{1} + \frac{1}{2} - \frac{1}{(n+1)} - \frac{1}{(n+2)})$$

We can use similar concept to such kind of problems.

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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/  [#permalink]

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01 Aug 2015, 23:45
iamheisenberg wrote:
Just want to post a general strategy to solve this kind of question. As this is already solved, I will try on another similar kind. Let's try to find out the value of the below expression:

$$\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}$$

The key to solve these type of questions is to rewrite each term as difference of two fractions and then cancel out maximum possible terms.

How can we rewrite each term as difference of two fractions??

1st term : $$\frac{1}{1*3} = (\frac{1}{2})\frac{2}{1*3} = (\frac{1}{2})\frac{3-1}{1*3} = (\frac{1}{2})(\frac{3}{1*3} - \frac{1}{1*3}) = (\frac{1}{2})(\frac{1}{1} - \frac{1}{3})$$

2nd term : $$\frac{1}{2*4} = (\frac{1}{2})\frac{2}{2*4} = (\frac{1}{2})\frac{4-2}{2*4} = (\frac{1}{2})(\frac{4}{2*4} - \frac{2}{2*4}) = (\frac{1}{2})(\frac{1}{2} - \frac{1}{4})$$

Similarly, 3rd term: $$\frac{1}{3*5} = (\frac{1}{2})(\frac{1}{3} - \frac{1}{5})$$

(n-1)th term: $$\frac{1}{(n-1)*(n+1)} = (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)})$$

nth term: $$\frac{1}{n*(n+2)} = (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})$$

Now using the above terms in the expression:

$$\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}$$

= $$(\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) +(\frac{1}{2})(\frac{1}{4} - \frac{1}{6}) + ........ + (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)}) + (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})$$

= $$(\frac{1}{2}) (\frac{1}{1}$$ $$- \frac{1}{3}$$ $$+ \frac{1}{2}$$ $$- \frac{1}{4}$$ $$+ \frac{1}{3}$$ $$- \frac{1}{5}$$ + $$\frac{1}{4}$$ $$- \frac{1}{6} + ....... + \frac{1}{(n-2)}$$ $$- \frac{1}{n}$$ + $$\frac{1}{(n-1)} - \frac{1}{(n+1)}$$ $$+ \frac{1}{n}$$ $$- \frac{1}{(n+2)})$$

In the above expression, you can notice that the same coloured terms can be cancelled out.

Now after cancelling we can simplify the complete expression to

$$(\frac{1}{2}) (\frac{1}{1} + \frac{1}{2} - \frac{1}{(n+1)} - \frac{1}{(n+2)})$$

We can use similar concept to such kind of problems.

I have doubt about it!
Ain't it time consuming and how much time did you consumed to solve this math your way? ( Just curious so would be better if you mention time needed to solve your way of thinking. I appreaciate thinking about possibilities but if during GMAT exam we are under time constrain so students always go after those techniques that helps them to pick the right answer in best possible time.

Thanks.
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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/  [#permalink]

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02 Aug 2015, 03:07
sashibagra wrote:
iamheisenberg wrote:
Just want to post a general strategy to solve this kind of question. As this is already solved, I will try on another similar kind. Let's try to find out the value of the below expression:

$$\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}$$

The key to solve these type of questions is to rewrite each term as difference of two fractions and then cancel out maximum possible terms.

How can we rewrite each term as difference of two fractions??

1st term : $$\frac{1}{1*3} = (\frac{1}{2})\frac{2}{1*3} = (\frac{1}{2})\frac{3-1}{1*3} = (\frac{1}{2})(\frac{3}{1*3} - \frac{1}{1*3}) = (\frac{1}{2})(\frac{1}{1} - \frac{1}{3})$$

2nd term : $$\frac{1}{2*4} = (\frac{1}{2})\frac{2}{2*4} = (\frac{1}{2})\frac{4-2}{2*4} = (\frac{1}{2})(\frac{4}{2*4} - \frac{2}{2*4}) = (\frac{1}{2})(\frac{1}{2} - \frac{1}{4})$$

Similarly, 3rd term: $$\frac{1}{3*5} = (\frac{1}{2})(\frac{1}{3} - \frac{1}{5})$$

(n-1)th term: $$\frac{1}{(n-1)*(n+1)} = (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)})$$

nth term: $$\frac{1}{n*(n+2)} = (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})$$

Now using the above terms in the expression:

$$\frac{1}{1*3} + \frac{1}{2*4} + \frac{1}{3*5} + \frac{1}{4*6} + \frac{1}{5*7} + ........ + \frac{1}{(n-1)*(n+1)} + \frac{1}{n*(n+2)}$$

= $$(\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) +(\frac{1}{2})(\frac{1}{4} - \frac{1}{6}) + ........ + (\frac{1}{2})(\frac{1}{(n-1)} - \frac{1}{(n+1)}) + (\frac{1}{2})(\frac{1}{n} - \frac{1}{(n+2)})$$

= $$(\frac{1}{2}) (\frac{1}{1}$$ $$- \frac{1}{3}$$ $$+ \frac{1}{2}$$ $$- \frac{1}{4}$$ $$+ \frac{1}{3}$$ $$- \frac{1}{5}$$ + $$\frac{1}{4}$$ $$- \frac{1}{6} + ....... + \frac{1}{(n-2)}$$ $$- \frac{1}{n}$$ + $$\frac{1}{(n-1)} - \frac{1}{(n+1)}$$ $$+ \frac{1}{n}$$ $$- \frac{1}{(n+2)})$$

In the above expression, you can notice that the same coloured terms can be cancelled out.

Now after cancelling we can simplify the complete expression to

$$(\frac{1}{2}) (\frac{1}{1} + \frac{1}{2} - \frac{1}{(n+1)} - \frac{1}{(n+2)})$$

We can use similar concept to such kind of problems.

I have doubt about it!
Ain't it time consuming and how much time did you consumed to solve this math your way? ( Just curious so would be better if you mention time needed to solve your way of thinking. I appreaciate thinking about possibilities but if during GMAT exam we are under time constrain so students always go after those techniques that helps them to pick the right answer in best possible time.

Thanks.

Not at all. It didn't take much time - probably a minute or less. It's just the way I explained it in the question is lengthy. Otherwise, once you get this concept, I think it is much easier, quicker and more accurate to do this way than the alternative way of calculating each term's approx value and summing it up.

Actually I had solved similar problems many years back in school and I knew the concept. So as soon as I saw this question I knew how to approach. May be that's why it took only a minute or less for me. And that is the reason I wanted to share the approach to such kind of questions (not only to this question).

But there are always different methods to solve a problem. And other people may find it easier to calculate the value of each term, sum them and check the answer choices.
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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/  [#permalink]

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02 Aug 2015, 04:12
sashibagra wrote:

I have doubt about it!
Ain't it time consuming and how much time did you consumed to solve this math your way? ( Just curious so would be better if you mention time needed to solve your way of thinking. I appreaciate thinking about possibilities but if during GMAT exam we are under time constrain so students always go after those techniques that helps them to pick the right answer in best possible time.

Thanks.

Not really. If you end up going for longer methods in solving GMAT questions, then you have surely missed out on certain information or methods that could have saved you some useful seconds (every second counts in GMAT!). You can see it this way: if lets say after looking at this question, your 1st instinct was to solve it by the fraction adding method (by taking the common LCM etc) or your 1st instinct was to solve it by converting fractions to decimals, then surely you have missed something. GMAT quant questions always have elegant and less time intensive solutions.

I do agree that this method of writing the given sequence in a particular way wont come straightaway but analyzing a particular question to see whether there is a method that will save time will help you in gaining those useful extra seconds. For me, I looked at the question and gave it 5-10 seconds as to how to tackle it. Converting to fractions was very straightforward and thus time consuming. It took me another 15 seconds to figure out what could be the relation (1/an+1 - 1/an etc) and finally another 20-25 seconds to write down the given sequence in the most productive manner. In total the total time was ~1 minute.

If you wouldve gone the fraction to decimal route, you would have spent some crucial seconds in calculating whats the decimal representation of 1/42 or 1/56 etc.
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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/  [#permalink]

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03 Aug 2015, 13:01
RIght off the bat, you can see that 1/2 + 1/4 = 3/4. So, the answer is already greater that choices A), B) or C). Based on the acscending nature of the answer choices, we can conclude that 46/55 is less than 9/10. But just using some reasoning, we are already at .75, 1/90 is a bit more than .01, 1/30 is a bit more than .03, 1/20 is .05. Adding these into .75, we are already at .84. Knowing the other 3 terms will give us approximately another .05, we can safely choose answer choice 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/  [#permalink]

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09 Jan 2019, 06:25
why is it wrong to calculate the sum by multiplying the mean (1/5*6) with the total number of items (9)?

That way we get 3/10 which is incorrect.
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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/  [#permalink]

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09 Jan 2019, 06:57
Sum = mean * number of terms

The mean (1/5*6) quoted by you is incorrect.

nausherwan wrote:
why is it wrong to calculate the sum by multiplying the mean (1/5*6) with the total number of items (9)?

That way we get 3/10 which is incorrect.

Posted from my mobile device
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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/  [#permalink]

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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/   [#permalink] 10 Jan 2020, 01:46
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