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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)(

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16 Apr 2016, 10:57
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8
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Difficulty:

65% (hard)

Question Stats:

56% (01:21) correct 44% (01:55) wrong based on 179 sessions

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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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16 Apr 2016, 11:17
1
6
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. $$\frac{9}{10}$$

We have 1/n(n+1) = 1/n - 1/(n+1)
So, $$\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\frac{1}{(5)(6)}+\frac{1}{(6)(7)}+\frac{1}{(7)(8)}+\frac{1}{(8)(9)}+\frac{1}{(9)(10)} = 1- \frac{1}{10} = \frac{9}{10}$$.
##### General Discussion
VP
Joined: 07 Dec 2014
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16 Apr 2016, 16:38
1
2
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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14 Nov 2016, 09:08
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

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Joined: 17 Aug 2016
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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. $$\frac{9}{10}$$

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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Joined: 08 Oct 2016
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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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17 Nov 2016, 15:15
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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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

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.

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28 Nov 2016, 15:41
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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25 Jul 2018, 22:07
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)( &nbs [#permalink] 25 Jul 2018, 22:07
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