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

Question

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}

anhnguyen178 · Accepted Answer

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} .

SSM700plus · Answer

.

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.

gracie · Answer

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

mvictor · Answer

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.

bazu · Answer

Could you please elaborate more on that? I can't understand wha you have applied

gillyelephant · Answer

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..

law258 · Answer

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

mvictor · Answer

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.

stonecold · Answer

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

Kinshook · Answer

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

AcetheGMAT2019 · Answer

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

___
Please give kudos if you find this helpful

TheNightKing · Answer

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.

dave13 · Answer

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

GMATinsight · Answer

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!

dave13 · Answer

many thanks GMATinsight but i still have question

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

GMATinsight · Answer

EXPERIENCE and EXPOSURE

Next time you too will know...

KarishmaB · Answer

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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