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The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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13 Dec 2010, 07:59
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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
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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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13 Dec 2010, 08:09
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



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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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13 Dec 2010, 08:24
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mmcooley33 wrote: 3/4, 5/36, 7/144 The numbers above form a sequence,t1,t2, and t3 , which is defined by tm = (1 / m^2)  (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 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.
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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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13 Dec 2010, 08:36
These pattern questions always get me. Thanks for the explanation
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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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02 Apr 2013, 10:56
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Bunuel wrote: Question: is \(Sum_j=1\frac{1}{(j+1)^2}>\frac{63}{64}\) > is \((j+1)^2>64\)> is \(j>8\)?
Dear Bunuel Should it not be \((j+1)^2>64\)> is \(j>7\) Instead of Bunuel wrote: ]> is \(j>8\)? Thank you



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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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02 Apr 2013, 11:00



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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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05 Feb 2014, 07:38
Responding to a pm: Question: 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 Solution: The numbers given above don't help us in any calculations. We should try to write the sequence on our own. \(t_m = \frac{1}{m^2}  \frac{1}{(m+1)^2}\) \(t_1 = 1  1/4\) \(t_2 = 1/4  1/9\) \(t_3 = 1/9  1/16\) Notice that the second term cancels out the first term of the next number. So when we add all these numbers, we will be left with 1  the second term of the last number (because it will not get canceled) (1) j>8 The number of terms will be at least 9. The sum of first 9 terms \(= 1  \frac{1}{10^2} = 99/100\) This is greater than 63/64. As the number of terms keep increasing, the second term which is subtracted keeps getting smaller so the sum tends toward 1. Hence the sum will always be greater than 63/64. Sufficient. (2) j<16 The number of terms could be 1 or 9 or 15 etc If the number of terms is 1, the sum will be 3/4 which is less than 63/64. As discussed in statement 1, if the number of terms is 9, the sum will be greater than 63/64. Not sufficient. Answer (A)
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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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13 Sep 2014, 06:54
Can someone explain what the difference between this question here and this question is: thesequences1s2s3snissuchthatsn1n103947.htmlSeems like they kinda ask the same, but I dont understand why the final question here is: is (j+1)^2>64> is j>7 ? And in the other question we ask about the numerator, is k > 9 ? Is there any difference?



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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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16 Sep 2014, 20:56
Bunuel wrote: jainpiyushjain wrote: Bunuel wrote: Question: is \(Sum_j=1\frac{1}{(j+1)^2}>\frac{63}{64}\) > is \((j+1)^2>64\)> is \(j>8\)?
Dear Bunuel Should it not be \((j+1)^2>64\)> is \(j>7\) Instead of Bunuel wrote: ]> is \(j>8\)? Thank you Typo edited. Thank you. +1. I think j>7 is correct since J+1>8....So J>7, isn't it?



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Re: The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
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