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27 Jan 2022, 09:00
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:44) correct
40%
(02:31)
wrong
based on 169
sessions
History
Date
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\((1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16})(1 - \frac{1}{25})...(1 - \frac{1}{225}) = ?\)
A. \(\frac{1}{75}\)
B. \(\frac{8}{75}\)
C. \(\frac{8}{15}\)
D. \(\frac{8}{225}\)
E. \(\frac{1}{5}\)
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{1}{75}\)
B. \(\frac{8}{75}\)
C. \(\frac{8}{15}\)
D. \(\frac{8}{225}\)
E. \(\frac{1}{5}\)
Are You Up For the Challenge: 700 Level Questions
27 Jan 2022, 23:00
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Bunuel
\((1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16})(1 - \frac{1}{25})...(1 - \frac{1}{225}) = ?\)
The most critical point is to understand that you are dealing with squares 4, 9, 16 and so on. Convert them accordingly.
\((1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{4^2})(1 - \frac{1}{5^2})...(1 - \frac{1}{15^2}) = ?\)
From here on you can do it in two ways.
(I)
\((\frac{2^2-1}{2^2})(\frac{3^2-1}{3^2})(\frac{4^2-1}{4^2})(\frac{5^2-1}{5^2})...(\frac{15^2-1}{15^2}) = ?\)
\((\frac{(2-1)(2+1)}{2^2})(\frac{(3-1)(3+1)}{3^2})( \frac{(4-1)(4+1)}{4^2})(\frac{(5-1)(5+1)}{5^2})...(\frac{(15-1)(15+1)}{15^2}) = ?\)
\(\frac{1*3*2*4*3*5*4*6…..12*14*13*15*14*16}{2^2*3^2*…….15^2}\)
\(\frac{1*2*3^2*4^2*5^2….13^2*14^2*15*16}{2^2*3^2*…15^2}\)
\(\frac{16}{2*15}=\frac{8}{15}\)
(II) \((1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{4^2})(1 - \frac{1}{5^2})...(1 - \frac{1}{15^2}) = ?\)
\((1-\frac{1}{2})(1+\frac{1}{2})* (1-\frac{1}{3})(1+\frac{1}{3})* (1-\frac{1}{4})(1+\frac{1}{4})* (1-\frac{1}{5})(1+\frac{1}{5})*……….. (1-\frac{1}{14})(1+\frac{1}{14})* (1-\frac{1}{15})(1+\frac{1}{15})\)
\(\frac{1}{2}*\frac{3}{2}*\frac{2}{3}*\frac{4}{3}*\frac{3}{4}*\frac{5}{4}*…….\frac{13}{14}*\frac{15}{14}*\frac{14}{15}*\frac{16}{15}\)
All terms except first and the last will get cancelled.
\(\frac{1}{2}*\frac{16}{15}=\frac{8}{15}\)
(III) You can also do it directly without getting into squares.
\((1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16})(1 - \frac{1}{25})...(1 - \frac{1}{225}) = ?\)
\(\frac{3}{4}*\frac{8}{9}*\frac{15}{16}*\frac{24}{25}*…….\frac{224}{225}\)
\(\frac{3*1}{2*2}*\frac{4*2}{3*3}*\frac{5*3}{4*4}*\frac{6*4}{5*5}*…….\frac{14*16}{225}\)
\(\frac{1}{2}*\frac{3}{2}*\frac{2}{3}*\frac{4}{3}*\frac{3}{4}*\frac{5}{4}*…….\frac{13}{14}*\frac{15}{14}*\frac{14}{15}*\frac{16}{15}\)
\(\frac{1}{2}*\frac{16}{15}=\frac{8}{15}\)
C
General Discussion
28 Jan 2022, 06:00
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If you simplify the first few terms you will see this pattern:
((3*1)/2^2)*((4*2)/3^2)*((5*3)/4^2))*((6*4)/5^2)
in which a factor in the numerator is repeated such that it cancels the square of that factor in the denominator, leaving 1,except for the first and last terms, which reduces to:
(1/2)*1*1*1*...(16/15)
= 8/15
Posted from my mobile device
((3*1)/2^2)*((4*2)/3^2)*((5*3)/4^2))*((6*4)/5^2)
in which a factor in the numerator is repeated such that it cancels the square of that factor in the denominator, leaving 1,except for the first and last terms, which reduces to:
(1/2)*1*1*1*...(16/15)
= 8/15
Posted from my mobile device
