I bet on option a
even by quick scan of answers it's clear the answer should be -ve
S(1)=-1/2, S(2)=1/6
S(3)=-1/12, S(4)=1/20
in total 5 such pairs which should return -ve value, as each successive element of the set is less than the preceding one (-ve value).
Tough One...Any Takers?
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shankar.ashwin
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There's no such pattern here to generalize, so list out the first few number.
Some general observations
- signs alternate
- magnitude gets smaller as n increases.
S1 = -1/2 = -0.5
S2 = 1/6 = 0.166
S3 = -1/12 = -0.083
S4 = 1/20 = 0.05
Now stop here, isolate S1, we notice (S2+S3+S4) is +ve and when added with S1 would result in a number slightly greater than -0.5 (-0.5 - 0). Only option B is in that range.
PS: Same pattern would follow until S10, isolating S1 we have 5 +ve numbers and 4 -ve with S2 being the highest magnitude, (S2+S3+----+S10) would be +ve, so S1(-0.5)+ (+ve number) would result in number > -0.5
Some general observations
- signs alternate
- magnitude gets smaller as n increases.
S1 = -1/2 = -0.5
S2 = 1/6 = 0.166
S3 = -1/12 = -0.083
S4 = 1/20 = 0.05
Now stop here, isolate S1, we notice (S2+S3+S4) is +ve and when added with S1 would result in a number slightly greater than -0.5 (-0.5 - 0). Only option B is in that range.
PS: Same pattern would follow until S10, isolating S1 we have 5 +ve numbers and 4 -ve with S2 being the highest magnitude, (S2+S3+----+S10) would be +ve, so S1(-0.5)+ (+ve number) would result in number > -0.5
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[email protected]
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We can sum the series upto S10 as:
-1/2 +(1/6-1/12) + (1/20-1/30) + (1/42-1/56) + (1/72-1/90) + 1/110
=-1/2 + (1/12)+ (1/60) + (1/168) + (1/360)+ 1/110
=-1/2 + (+ve fraction)
So definitely Sum of first 10 member of this series would be slightly > -1/2
Hence the best answer choice would be B.
-1/2 +(1/6-1/12) + (1/20-1/30) + (1/42-1/56) + (1/72-1/90) + 1/110
=-1/2 + (1/12)+ (1/60) + (1/168) + (1/360)+ 1/110
=-1/2 + (+ve fraction)
So definitely Sum of first 10 member of this series would be slightly > -1/2
Hence the best answer choice would be B.
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ronnie1985
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The question is not appearing in full... please post the full question...
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pemdas
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damned it, i could not read the sign negative 1/2 and 0 in answer choice B and decided right away about the only negative in a)
it's b if one tests the sum of resultants
it's b if one tests the sum of resultants
pemdas wrote:I bet on option a
even by quick scan of answers it's clear the answer should be -ve
S(1)=-1/2, S(2)=1/6
S(3)=-1/12, S(4)=1/20
in total 5 such pairs which should return -ve value, as each successive element of the set is less than the preceding one (-ve value).
Success doesn't come overnight!
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GMATGuruNY
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Calculate just enough to see the pattern.
If n=1, (-1)¹ * 1/(1*2) = -1/2.
If n=2, (-1)² * 1/(2*3) = 1/6
If n=3, (-1)³ * 1/(3*4) = -1/12.
If n=4, (-1)� * 1/(4*5) = 1/20.
Plot the results on a number line:
The sum of the first two terms = -1/2 + 1/6 = -1/3.
From this point on, the sum alternates between decreasing by a very small fraction and increasing by an even smaller fraction.
Thus, the sum will end up somewhere between -1/2 (the first term) and -1/3 (the sum of the first two terms).
The correct answer is B.
If n=1, (-1)¹ * 1/(1*2) = -1/2.
If n=2, (-1)² * 1/(2*3) = 1/6
If n=3, (-1)³ * 1/(3*4) = -1/12.
If n=4, (-1)� * 1/(4*5) = 1/20.
Plot the results on a number line:
The sum of the first two terms = -1/2 + 1/6 = -1/3.
From this point on, the sum alternates between decreasing by a very small fraction and increasing by an even smaller fraction.
Thus, the sum will end up somewhere between -1/2 (the first term) and -1/3 (the sum of the first two terms).
The correct answer is B.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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