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The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {
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04 Oct 2013, 02:09
04 Oct 2013, 02:09
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Difficulty:
45%
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Question Stats:
74% (02:48) correct
26%
(02:35)
wrong
based on 238
sessions
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
A) 5/12
B) 1/2
C) 11/20
D) 2/3
E) 7/10
The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {
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04 Oct 2013, 02:22
04 Oct 2013, 02:22
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Expert Reply
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
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.
General Discussion
The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - {
[#permalink]
04 Oct 2013, 02:22
04 Oct 2013, 02:22
3
Kudos
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
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.
