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13 Dec 2010, 07:59
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A
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Difficulty:
75%
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Question Stats:
61% (02:43) correct
39%
(02:31)
wrong
based on 464
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3/4, 5/36, 7/144
The numbers above form a sequence, t1, t2, and t3 , which is defined by \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\) for all positive integers m. Is the sum of the first J terms of the sequence greater than 63/64?
(1) j>8
(2) j<16
The numbers above form a sequence, t1, t2, and t3 , which is defined by \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\) for all positive integers m. Is the sum of the first J terms of the sequence greater than 63/64?
(1) j>8
(2) j<16
13 Dec 2010, 08:24
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mmcooley33
In such kind of questions there is always a pattern in terms or/and in the sum of the terms.
Given: \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\). So:
\(t_1=\frac{1}{1^2}-\frac{1}{(1+1)^2}=1-\frac{1}{2^2}\);
\(t_2=\frac{1}{2^2}-\frac{1}{(2+1)^2}=\frac{1}{2^2}-\frac{1}{3^2}\);
\(t_3=\frac{1}{3^2}-\frac{1}{(3+1)^2}=\frac{1}{3^2}-\frac{1}{4^2}\);
...
You should notice that if we have as sum of first 2 terms then every thing but the 1 from \(t_1\) and the last part from \(t_2\) (1/3^2=1/(2+1)^2) will cancel out, so \(sum_2=1-\frac{1}{(2+1)^2}\). The same if we sum first 3 terms: only 1 minus the last part of \(t_3\) (1/4^2=1/(3+1)^2) will remain, \(sum_3=1-\frac{1}{(3+1)^2}\). So if we sum first \(j\) terms the the sum will equal to \(1-\frac{1}{(j+1)^2}\).
Question: is \(Sum_j=1-\frac{1}{(j+1)^2}>\frac{63}{64}\) --> is \((j+1)^2>64\)--> is \(j>7\)?
(1) j>8. Sufficient.
(2) j<16. Not sufficient.
Answer: A.
General Discussion
13 Dec 2010, 08:09
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the solution has to do with manipulation of the negative. The answer given by knewton turns 1/m^2 - 1/((m+1)^2) into (-1/M^2 + 1/m^2) so i get how it turns all but 1 into zero but they say the last term (-1/j^2 + 1/j^2) - 1/((j+1)^2). I am probably missing a simple well known concept from order of operations but could someone please enlighten me. Plus if there is a better way to cut and paste formulas from websites please inform me thanks, otherwise i would post their solution for reference. Thanks a bunch
