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Sum of integers

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Joined: 23 Feb 2010
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Sum of integers [#permalink] New post 07 Mar 2010, 14:45
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (03:43) correct 0% (00:00) wrong based on 1 sessions
Please kindly provide explanations/shortcuts if possible. Thanks
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Re: Sum of integers [#permalink] New post 07 Mar 2010, 14:53
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gmatgg wrote:
Please kindly provide explanations/shortcuts if possible. Thanks


1.
If u notice, all rows starting from 2nd are mutliple of 1st row.
Sum of first row = 7*8/2 = 28 = k (lets say)

First row = k
Second row = -2k
third row = 3k
fourth row = -4k
fifth row = 5k
sixth row = -6k
seventh row = 7k

sum = k - 2k + 3k - 4k + 5k - 6k + 7k = 4k = 4*28 = 112 Ans
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Senior Manager
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Re: Sum of integers [#permalink] New post 07 Mar 2010, 15:02
gmatgg wrote:
Please kindly provide explanations/shortcuts if possible. Thanks


For 2nd ques, refer: sequence-88874.html?view-post=670848#p670848
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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
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Re: Sum of integers [#permalink] New post 07 Mar 2010, 16:21
Great explanation! thanks so much!!!!
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Re: Sum of integers [#permalink] New post 15 Mar 2010, 08:35
Problem 1
We can say sum of postivie integers are (1+3+5+7)(1+2+3+4+5+6+7) = 16*7*8/2 = 448 (sum of numbers from 1 to n is n(n+1)/2)
and the sum of all the negative numbers = (2+4+6)(1+2+3+4+5+6+7) = 12*7*8/2 = 336
Subtract second from first and answer is 112
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Re: Sum of integers [#permalink] New post 15 Mar 2010, 08:44
(-1)^k+1*(1/2^k)

(-1)^k+1 says that every odd number will be positive and even number will be negative
so series will be like following
1/2-1/2^2+1/2^3-1/2^4....

=1/2(1+1/4+1/8+1/16+1/32) - 1/4(1+1/4+1/8+1/16+1/32)
=(1/2 - 1/4)(1+1/4+1/8+1/16+1/32)
=1/4*(32+16+8+4+1)/32
=1/4*61/32 = 61/128 =
1/4 = 32/128 multiply 32 in n and d
and 1/2 is 64/128 multiply 64 in n and d

so correct answer is it is between 1/4 to 1/2
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Re: Sum of integers   [#permalink] 15 Mar 2010, 08:44
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