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The nth term in a certain sequence is defined for positive integer n a

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The nth term in a certain sequence is defined for positive integer n a  [#permalink]

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New post 22 Sep 2016, 04:41
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

59% (02:29) correct 41% (02:06) wrong based on 140 sessions

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Re: The nth term in a certain sequence is defined for positive integer n a  [#permalink]

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New post 22 Sep 2016, 05:08
1
sequence goes like this 0 , 1, 0 , 1, 0 ...... 1(64th term)

sum of 64 terms = 32

I go with D
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Re: The nth term in a certain sequence is defined for positive integer n a  [#permalink]

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New post 29 Oct 2016, 03:02
deepthit wrote:
sequence goes like this 0 , 1, 0 , 1, 0 ...... 1(64th term)

sum of 64 terms = 32

I go with D

Could you, please, elaborate your explanation?
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Re: The nth term in a certain sequence is defined for positive integer n a  [#permalink]

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New post 29 Oct 2016, 07:38
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Bunuel wrote:
The nth term in a certain sequence is defined for positive integer n as \(\frac{1+(-1)^{(\frac{n(n+1)}{2})}}{2}\). What is the total of the first 64 terms of this sequence?

A. −164
B. 0
C. 16
D. 32
E. 64


To find term1, replace n with 1 in the expression \(\frac{1+(-1)^{(\frac{n(n+1)}{2})}}{2}\)
We find that term1 = 0

To find term2, replace n with 2 in the expression \(\frac{1+(-1)^{(\frac{n(n+1)}{2})}}{2}\)
We find that term2 = 0

To find term3, replace n with 3 in the expression. We find that term3 = 1

To find term4, replace n with 4 in the expression. We find that term4 = 1

Term5 = 0

Term6 = 0

Term7 = 1

Term8 = 1
.
.
.

So, the SUM of the first 64 terms of the sequence = 0 + 0 + 1 + 1 + 0 + 0 + 1 + 1 .....

As we can see, HALF of the terms will be 0 and HALF will be 1
In other words, 32 terms will equal 0 and 32 terms will equal 1
So, the sum = (32)(0) + (32)(1) = 32

Answer:

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Re: The nth term in a certain sequence is defined for positive integer n a  [#permalink]

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Re: The nth term in a certain sequence is defined for positive integer n a   [#permalink] 06 Aug 2018, 03:50
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