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The numbers above form a sequence, t1, t2, and t3 , which is [#permalink]
13 Dec 2010, 06:59

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Difficulty:

65% (hard)

Question Stats:

67% (04:07) correct
33% (02:02) wrong based on 141 sessions

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?

Re: Knewton Diagnostic Question #28 [#permalink]
13 Dec 2010, 07:09

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

Re: Knewton Diagnostic Question #28 [#permalink]
13 Dec 2010, 07:24

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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:

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 \(j>7\)?

The numbers above form a sequence, t1, t2, and t3 [#permalink]
05 Feb 2014, 06:38

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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.

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