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# The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {

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Director
Joined: 29 Nov 2012
Posts: 774
The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {  [#permalink]

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04 Oct 2013, 02:09
4
1
6
00:00

Difficulty:

55% (hard)

Question Stats:

72% (02:50) correct 28% (02:29) wrong based on 194 sessions

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The product $$(1 - {\frac{1}{2}}^2)(1 - {\frac{1}{3}}^2)...(1 - {\frac{1}{9}}^2)(1 - {\frac{1}{10}}^2)$$ equals...

A) 5/12
B) 1/2
C) 11/20
D) 2/3
E) 7/10

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Joined: 02 Sep 2009
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The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {  [#permalink]

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04 Oct 2013, 02:22
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4
fozzzy wrote:
The product $$(1 - {\frac{1}{2}}^2)(1 - {\frac{1}{3}}^2)...(1 - {\frac{1}{9}}^2)(1 - {\frac{1}{10}}^2)$$ equals...

A) 5/12
B) 1/2
C) 11/20
D) 2/3
E) 7/10

$$(1 - {\frac{1}{2}}^2)(1 - {\frac{1}{3}}^2)...(1 - {\frac{1}{9}}^2)(1 - {\frac{1}{10}}^2)=$$

$$=(1-\frac{1}{2})(1+\frac{1}{2})(1-\frac{1}{3})(1+\frac{1}{3})...(1-\frac{1}{9})(1+\frac{1}{9})(1-\frac{1}{10})(1+\frac{1}{10})=$$

$$=\frac{1}{2}*(\frac{3}{2}*\frac{2}{3})*(\frac{4}{3}*\frac{3}{4})*(\frac{5}{4}*\frac{4}{5})*(\frac{6}{5}*...)...(...\frac{8}{9})*(\frac{10}{9}*\frac{9}{10})*\frac{11}{10}=\frac{1}{2}*\frac{11}{10}=\frac{11}{20}$$

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##### General Discussion
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 613
The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {  [#permalink]

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04 Oct 2013, 02:22
3
fozzzy wrote:
The product $$(1 - {\frac{1}{2}}^2)(1 - {\frac{1}{3}}^2)...(1 - {\frac{1}{9}}^2)(1 - {\frac{1}{10}}^2)$$ equals...

A) 5/12
B) 1/2
C) 11/20
D) 2/3
E) 7/10

Break $$(1 - {\frac{1}{2}}^2)$$ = $$(1 + {\frac{1}{2}})*(1 - {\frac{1}{2}})$$
Similarly, the given product will be of two parts :$$(1 + {\frac{1}{2}})*(1 + {\frac{1}{3}}).....(1 + {\frac{1}{10}}) = \frac{3}{2}*\frac{4}{3}*\frac{5}{4}....\frac{11}{10}$$

Notice that all the terms will cancel in alternate expressions except 2 in the denominator for the first term and 11 in the numerator for the last term. Also, as the other set which is: $$(1 - {\frac{1}{2}})*(1 - {\frac{1}{3}}).....(1 - {\frac{1}{10}})$$ can never have an 11 in the denominator, the correct answer will be something which has 11 in the numerator : C

C.
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Joined: 30 Mar 2013
Posts: 109
The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {  [#permalink]

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06 Oct 2014, 12:03
My way was the long way, but the numerators can all be written as 1 less than the denominator, and the denominator is square of every number from 2 till 10:

3/4 * 8/9 * 15/16 * 24/25 * 35/36* 48/49 * 63/64* 80/81 *99/100

Cancelling out everything leaves 11/20. Took me around 2 mins though.
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Joined: 09 Sep 2013
Posts: 8543
Re: The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {  [#permalink]

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22 Oct 2017, 08:54
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Re: The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { &nbs [#permalink] 22 Oct 2017, 08:54
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