Given a series n_k = ((-1)^k)*k, such that k is a positive : DS Archive
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# Given a series n_k = ((-1)^k)*k, such that k is a positive

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Given a series n_k = ((-1)^k)*k, such that k is a positive [#permalink]

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27 May 2005, 19:49
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Given a series n_k = ((-1)^k)*k, such that k is a positive integer from 1 to N, is N odd?

1) Sum of all elemets of the series is less than N.
2) Sum of all elemets of the series is a negative integer.
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28 May 2005, 02:24
The series is

-1, 2, -3, 4, -5, 6, ....

This seq till every even number can be considered in groups of two, each of which has a sum 1.

Thus, S(2) = 1, S(4) = 2, s(6) = 3 and so on.

But this seq's sum till every odd number is (seq no -1) /2 - the odd number.

Thus, S(3) = 1 -3 = -2, S(5) = 2 - 5 = -3 and so on.

From 1), Sum of all elements of the series is less than N.
This is true in all cases. Doesnt tell whether N is even or odd.

From 2), N has to be odd (see the above illustrations). For N even the sum is always + N/2, but is always -ve for odd N.

Thus B.
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28 May 2005, 05:26
I too vote for B

I have a simple explanation, the series is
( -1 + 2 ) + ( -3 + 4 ) .....to N terms.

Clearly is N is even then the sum is 1*No of pairs = n/2
If N is odd then the sum is (sum till the n-1 th term) + nth term
= (n-1)/2 - n = -(n+1)/2

1) is true for both even and odd n
2) is true only when n is odd

Hence B

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28 May 2005, 10:11
B for me too

I) N can be even or odd...we will notice that elements in the series alternate signs...thus the sum of all elements can be less than N but N can be even or odd....

II) N has to be odd in order for the elements to be negative....

B it s
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28 May 2005, 13:26
sparky wrote:
Given a series n_k = ((-1)^k)*k, such that k is a positive integer from 1 to N, is N odd?

1) Sum of all elemets of the series is less than N.
2) Sum of all elemets of the series is a negative integer.

B is the OA

N is odd => sum = (N-1)/2 - N < 0
N is even => sum = 0< N/2 <N

1) insufficient
2) sufficient
Re: DS II   [#permalink] 28 May 2005, 13:26
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