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What is (-1)^1 + (-1)^2 + ... + (-1)^2006?

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Joined: 02 Sep 2009
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What is (-1)^1 + (-1)^2 + ... + (-1)^2006?  [#permalink]

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26 Mar 2019, 05:18
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Difficulty:

5% (low)

Question Stats:

83% (00:34) correct 17% (00:39) wrong based on 52 sessions

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What is $$(-1)^1 + (-1)^2 + ... + (-1)^{2006}$$?

(A) -2006
(B) -1
(C) 0
(D) 1
(E) 2006

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Re: What is (-1)^1 + (-1)^2 + ... + (-1)^2006?  [#permalink]

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26 Mar 2019, 05:27
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If the power of the end term is odd then -1 and if even then 0. As 2006 is even it is option c 0

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Re: What is (-1)^1 + (-1)^2 + ... + (-1)^2006?  [#permalink]

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26 Mar 2019, 07:00
Top Contributor
Bunuel wrote:
What is $$(-1)^1 + (-1)^2 + ... + (-1)^{2006}$$?

(A) -2006
(B) -1
(C) 0
(D) 1
(E) 2006

In TOTAL, there are 2006 terms to add
The pattern alternates between -1 and 1: (-1) + 1 + (-1) + 1 + (-1) + 1 + . . . . . . (-1) + 1
Every PAIR of consecutive values add to 0
That is, (-1) + 1 = 0

So, (-1) + 1 + (-1) + 1 + (-1) + 1 + . . . . . . (-1) + 1 = 0 + 0 + 0 . . . . + 0 + 0
Since we have an EVEN number of terms (2006 terms), we have exactly 1003 PAIRS (of -1 and 1) that add to 0
So, the sum becomes the sum of 1003 0's, which is ZERO

Cheers,
Brent
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Re: What is (-1)^1 + (-1)^2 + ... + (-1)^2006?  [#permalink]

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26 Mar 2019, 11:13
Bunuel wrote:
What is $$(-1)^1 + (-1)^2 + ... + (-1)^{2006}$$?

(A) -2006
(B) -1
(C) 0
(D) 1
(E) 2006

pair of odd & even ; 2006/2 ; 1003
so equal of both sets sum = 0
IMO C
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Re: What is (-1)^1 + (-1)^2 + ... + (-1)^2006?   [#permalink] 26 Mar 2019, 11:13
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