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

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fozzzy
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The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { [#permalink] New post   04 Oct 2013, 02:09
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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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C
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Bunuel
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The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { [#permalink] New post   04 Oct 2013, 02:22
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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


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

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

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

Answer: C.
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mau5
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The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { [#permalink] New post   04 Oct 2013, 02:22
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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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usre123
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The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { [#permalink] New post   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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Re: The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { [#permalink] New post   15 Mar 2022, 18:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { [#permalink]
15 Mar 2022, 18:09
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