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# What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +....

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What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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18 Mar 2014, 14:59
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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) + 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.
[Reveal] Spoiler: OA

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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18 Mar 2014, 22:53
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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) +.... [#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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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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19 Mar 2014, 18:55
GyanOne wrote:
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).

This is excellent. How long it took to derive the formula?
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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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17 Jun 2015, 23:15
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VeritasPrepKarishma:

Can you please suggest a efficient method to solve this.

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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17 Jun 2015, 23:36
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Shree9975 wrote:
:

Can you please suggest a efficient method to solve this.

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) +.... [#permalink]

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17 Jun 2015, 23:42
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Shree9975 wrote:

Can you please suggest a efficient method to solve this.

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)

Add Them up

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

Answer: Option
[Reveal] Spoiler:
E

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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18 Jun 2015, 01:05
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thanks GMATinsight.
It helped.

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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18 Jun 2015, 01:09
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thanks GMATinsight.
It helped.

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#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.

Is this helpful?

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What is 1/(1)(2)+1/(2)(3) [#permalink]

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31 Jul 2015, 08:55
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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

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Last edited by ENGRTOMBA2018 on 31 Jul 2015, 09:08, edited 1 time in total.
Formatted the question and edited the topic

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What is 1/(1)(2)+1/(2)(3) [#permalink]

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31 Jul 2015, 09:22
Gmat1008 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)2/5
b)3/5
c)7/10
d)46/55
e)9/10

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Please format your question properly and help GMATCLUB to maintain the forums in a proper manner.

Note that the elements of the sequence can be written in the form: $$a_n = \frac{1}{a_n * (a_{n+1})}$$ = $$\frac{1}{a_n} - \frac{1}{a_{n+1}}$$

Thus the sequence can be written as : $$\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3}+ \frac{1}{3} - \frac{1}{4}+\frac{1}{4} - \frac{1}{5}+\frac{1}{5} - \frac{1}{6}+ \frac{1}{6} - \frac{1}{7}+\frac{1}{7} - \frac{1}{8}+\frac{1}{8} - \frac{1}{9}+\frac{1}{9} - \frac{1}{10} = 1- \frac{1}{10} = \frac{9}{10}$$ .

E is the correct answer.

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What is 1/(1)(2)+1/(2)(3) [#permalink]

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01 Aug 2015, 04:03
Hey,

I didn't think of seeing it as a sequence. What I did is that I tried to find a way to simplify the result - round them off.

So, I started with calculating the first two fractions, ending up with 8/12.
8/12 * 1/12 = 9/12. Then 9/12 * 1/20 = 192/240. Now, both the numberator and the denominator are close to 200. I decided that this would be close to 1.

Then, adding the rest of the fractions always leads to sth like 21/20, with the numerator being one more than the denominator. So, the final result should be close to 1, and E is close to 1.

However, it took about 3 1/2 minutes to do it.

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Re: What is 1/(1)(2)+1/(2)(3) [#permalink]

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01 Aug 2015, 04:51
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Gmat1008 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)2/5
b)3/5
c)7/10
d)46/55
e)9/10

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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

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

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)

Add Them up

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

Answer: Option
[Reveal] Spoiler:
E

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Re: What is 1/(1)(2)+1/(2)(3) [#permalink]

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01 Aug 2015, 04:52
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Gmat1008 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)2/5
b)3/5
c)7/10
d)46/55
e)9/10

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Bunuel: The question has already been discussed here

what-is-168951.html
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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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01 Aug 2015, 05:40
GMATinsight wrote:
Gmat1008 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)2/5
b)3/5
c)7/10
d)46/55
e)9/10

Please help me unlock the awesome Gmatclub test by giving a Kudos

Bunuel: The question has already been discussed here

what-is-168951.html

Thanks for the observation, I have merged the topics. You can tag me or PM me as and when you find a question that had already been discussed on the forums. I usually merge topics as and when I find them. This one slipped under the radar.

Topics merged.

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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#permalink]

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01 Aug 2015, 05:50
Engr2012 wrote:
GMATinsight wrote:
Gmat1008 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)2/5
b)3/5
c)7/10
d)46/55
e)9/10

Please help me unlock the awesome Gmatclub test by giving a Kudos

Bunuel: The question has already been discussed here

what-is-168951.html

Thanks for the observation, I have merged the topics. You can tag me or PM me as and when you find a question that had already been discussed on the forums. I usually merge topics as and when I find them. This one slipped under the radar.

Topics merged.

There is always difference in tagging you and in tagging Bunuel because Bunuel Given Kudos for making such observation and you don't.

Anyways just kidding. Will keep in mind next time.
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Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#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.

Is this helpful?

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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Location: United Arab Emirates
Concentration: General Management, Operations
GMAT 1: 690 Q49 V34
GMAT 2: 730 Q50 V38
WE: Sales (Energy and Utilities)
Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +.... [#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.

Is this helpful?

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.

Kudos [?]: 57 [0], given: 57

Re: What is 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + /1(4)*(5) +....   [#permalink] 02 Aug 2015, 03:07

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