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For every integer k from 1 to 10, inclusive, the kth term of

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For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 14 Nov 2006, 15:55
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For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1)*(1/2^k). If T is the sum of the first 10 terms in the sequence then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 and 1
D. Between 1/4 and 1/2
E. Less than 1/4

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jul 2014, 10:52, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 14 Nov 2006, 16:52
(-1)^K+1 . (1/2^K). --> The dots mean multiply?
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 14 Nov 2006, 17:04
ywilfred wrote:
(-1)^K+1 . (1/2^K). --> The dots mean multiply?


Yes that is correct :-)
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 14 Nov 2006, 17:16
Q. For every integer K from 1 to 10, inclusive, the Kth term of a certain sequence is given by (-1)^K+1 . (1/2^K). If T is the sum of the first 10 terms in the sequence, then T is

a) > 2
b) between 1 and 2
c) between 1/2 and 1
d) between 1/4 and 1/2
e) less then 1/4


T1 = (-1)^2*(1/2)^1= 1/2
T2= (-1)^3*(1/2)^2 = -1/4
T3= (-1)^4*(1/2)^3 = + 1/8
.
.
T9= (-1)^10*(1/2)^9= (1/2)^9
t10=(-1)^11*(1/2)^10= - (1/2)^10

ans b/w 1/2 and 1

i think this problem is based on using the knowlege that fractions when multiplied by a fraction the value of product reduces

so 1/2 multiped by 1/2 is less than 1/2
and so on

since 1st term is 1/2 the sume should be greatd than 1/2

if u add rest of the trms the sume will b less than 1/2
ANS B between 1/2 and 1
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 14 Nov 2006, 17:30
I got D for this one. I used the same logic as Damager did. However, I think since the sequence alters a positive and a negative, ie the first term is 1/2, the second is -1/4, the third is 1/8, fourth is -1/16. So even with the first two terms, the sum should be 1/4 and then adding smaller and smaller amount as the sequence goes. So Asn D between 1/4 and 1/2
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 15 Nov 2006, 00:48
enola wrote:
I got D for this one. I used the same logic as Damager did. However, I think since the sequence alters a positive and a negative, ie the first term is 1/2, the second is -1/4, the third is 1/8, fourth is -1/16. So even with the first two terms, the sum should be 1/4 and then adding smaller and smaller amount as the sequence goes. So Asn D between 1/4 and 1/2


So adding smaller and smaller amounts makes it go to 1/2 even though the terms are positive and negative. Can you please explain a bit more about your logic?

Thanks
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 16 Nov 2006, 15:24
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The answer is D


--------------------------------------------------------------------------------

Q. For every integer K from 1 to 10, inclusive, the Kth term of a certain sequence is given by (-1)^K+1 . (1/2^K). If T is the sum of the first 10 terms in the sequence, then T is

a) > 2
b) between 1 and 2
c) between 1/2 and 1
d) between 1/4 and 1/2
e) less then 1/4


T1 = (-1)^2*(1/2)^1= 1/2
T2= (-1)^3*(1/2)^2 = -1/4
T3= (-1)^4*(1/2)^3 = + 1/8
.
.
T9= (-1)^10*(1/2)^9= (1/2)^9
t10=(-1)^11*(1/2)^10= - (1/2)^10

1/2 - 1/4 = 1/4 + 1/8 = 3/8 - 1/16 = 5/16 + 1/32 = 11 /32 - 1/64 = 21 / 64 + 1/128 = 43 / 128

since you are summing, the number is going to slowly move between 1/4 and 1/2 but staying above 1/4 and below 1/2. It keeps moving by smaller and smaller fractions, but it always will be above 1/4 and under 1/2. Look at it this way. The first two numbers establish the range. It starts at a 1/2 then drops to 1/4 and each time it lowers and raises by a smaller increment. What you will have is something that looks like this:

..
....
.......
........
...........
.............
.................
.............
...........
.........
.......
....
..

meaning, first the number swings a lot, then it slowly swings by increasingly an infinitely smaller increments. Eventually you are adding 1/10000000 and then subtracting 1/100000000 and so on into infinity.

Long story short, you have the number .5 and .25. The number made by this sum will just keep getting more and more decimal places instead of moving past one of those two numbers.

I really hope this long ass explanation is right or it will look pretty stupid :-).

I'm sitting in Norway and its almost 1am, so who knows.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 16 Nov 2006, 20:43
The series 1/2-1/4+1/8-1/16...is actually a geometric progression with the first term 1/2 and every subsequent term multiplied by -1/2.

a,ar,ar2,.....etc

Sum of 1st n terms of GP = a(1-r^[n+1])/(1-r)

Sum of 10 terms = 1/2(1-(-1/2)^[10+1])/(1-(-1/2) = 0.33

Answer : Between 1/4 and 1/2.

Hope this helps.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 24 Jul 2011, 17:39
ncp wrote:
The series 1/2-1/4+1/8-1/16...is actually a geometric progression with the first term 1/2 and every subsequent term multiplied by -1/2.

a,ar,ar2,.....etc

Sum of 1st n terms of GP = a(1-r^[n+1])/(1-r)

Sum of 10 terms = 1/2(1-(-1/2)^[10+1])/(1-(-1/2) = 0.33

Answer : Between 1/4 and 1/2.

Hope this helps.


Isnt the forumula for GP a(1-r^n)/1-r? See youtube video here : http://www.youtube.com/watch?v=OkPI1_BKo9w
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 25 Jul 2011, 03:30
ncp, what is short way to calculate (1+1/2^11) ? there is gotta be a shortcut..rather than calculating 2^11..
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 26 Jul 2014, 10:39
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 26 Jul 2014, 10:52
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Re: For every integer k from 1 to 10, inclusive, the kth term of   [#permalink] 26 Jul 2014, 10:52
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