Can someone please explain the following GMAT PREP question? I don't understand the language in the question? Please don't just answer the question  explain what the question is asking for.
For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)]*(1 / 2^k). If T is the sum of the first 10 terms of 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
OA is D
GMAT PREP QUESTION w/OA
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equation is [(1)^(k+1)]*(1 / 2^k)
for k = 1
you will get: [(1)^2]*(1/2) = 1*1/2=1/2
k=2 you will get (1)*1/4 = 1/4
k=3 you will get (1)* 1/8 = 1/8
k=4 you will get 1/16
k=5 you will get 1/32
and so on......
Note your series would look something like this
1/2  1/4 + 1/8  1/16 + 1/32..........
basically, you are subtracting 1/4 from 1/2 you get 1/4
then you add 1/8 to 1/4 = 3/8...
all in all, your end result for the series should be between 1/2 and 1/4
for k = 1
you will get: [(1)^2]*(1/2) = 1*1/2=1/2
k=2 you will get (1)*1/4 = 1/4
k=3 you will get (1)* 1/8 = 1/8
k=4 you will get 1/16
k=5 you will get 1/32
and so on......
Note your series would look something like this
1/2  1/4 + 1/8  1/16 + 1/32..........
basically, you are subtracting 1/4 from 1/2 you get 1/4
then you add 1/8 to 1/4 = 3/8...
all in all, your end result for the series should be between 1/2 and 1/4

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I don't even understand what the questions is asking for. The language is very odd. And how would you do your method in under 2 minutes?rohangupta83 wrote:equation is [(1)^(k+1)]*(1 / 2^k)
for k = 1
you will get: [(1)^2]*(1/2) = 1*1/2=1/2
k=2 you will get (1)*1/4 = 1/4
k=3 you will get (1)* 1/8 = 1/8
k=4 you will get 1/16
k=5 you will get 1/32
and so on......
Note your series would look something like this
1/2  1/4 + 1/8  1/16 + 1/32..........
basically, you are subtracting 1/4 from 1/2 you get 1/4
then you add 1/8 to 1/4 = 3/8...
all in all, your end result for the series should be between 1/2 and 1/4

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well, the question says that:
For any integer k from 1 to 10, inclusive  which means k could be 110 (inclusive)
kth of a certain sequence is given by [(1)^(k+1)]*(1 / 2^k)  which means that kth term is given by the above sequence
For example you want to find out the first term of the sequence (remember, there are only 10 terms in the sequence with k from 110) i.e. k=1
plug k = 1 into the above formual and you get the first term and so on
Finally  the question is asking you to find the sum of first 10 terms of this sequence i.e. find the sum of the sequence given by the formula [(1)^(k+1)]*(1 / 2^k) where 1<= k <=10
Hopefully this should clear what the question is asking you to do.
For any integer k from 1 to 10, inclusive  which means k could be 110 (inclusive)
kth of a certain sequence is given by [(1)^(k+1)]*(1 / 2^k)  which means that kth term is given by the above sequence
For example you want to find out the first term of the sequence (remember, there are only 10 terms in the sequence with k from 110) i.e. k=1
plug k = 1 into the above formual and you get the first term and so on
Finally  the question is asking you to find the sum of first 10 terms of this sequence i.e. find the sum of the sequence given by the formula [(1)^(k+1)]*(1 / 2^k) where 1<= k <=10
Hopefully this should clear what the question is asking you to do.
Last edited by rohangupta83 on Sun Oct 26, 2008 4:23 pm, edited 1 time in total.

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and in the solution, all you need are the first 4 terms at most to give you an idea what approximately is your answer.
For example
lets assume that the first term is 100
you take half out i.e subtract 50 so now you are left with 50
now you are adding 25 to this 50  you have 75
now subtracting 12.5 from 75  you have 62.5
and if you continue the same series which is adding half of what you just subtracted to the result and then subtracting half of what you just added from the result, you will notice that you will oscillate between the range of 50100, never crossing either extremes.
It is the same principle implied in the above problem
For example
lets assume that the first term is 100
you take half out i.e subtract 50 so now you are left with 50
now you are adding 25 to this 50  you have 75
now subtracting 12.5 from 75  you have 62.5
and if you continue the same series which is adding half of what you just subtracted to the result and then subtracting half of what you just added from the result, you will notice that you will oscillate between the range of 50100, never crossing either extremes.
It is the same principle implied in the above problem
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T = 1/2  1/2^2 + 1/2^3  ... 1/2^10
= 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5
Notice that 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 < 1/4, we can say that 1/4 < T < 1/2.
= 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5
Notice that 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 < 1/4, we can say that 1/4 < T < 1/2.
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Can someone please clarify how to solve this problem with 2 mins:
For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)]*(1 / 2^k). If T is the sum of the first 10 terms of the sequence, then T is:
For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)]*(1 / 2^k). If T is the sum of the first 10 terms of the sequence, then T is:
logitech wrote:This is an approximation question, hope this helps.
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For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)]*(1 / 2^k). If T is the sum of the first 10 terms of the sequence, then T is:
It has two parts (i) (1)^(k+1) we see that the answer will be +1 when k is odd and 1 when k is even
Why ? (1) ^even = +1 and (1)^odd = 1 , but when k = even k+1 will be odd and when k = odd k+1 will be even
(ii) so there will +ve values as well as ve value
From (i) we see that for even k the given function is ve and for odd k the given function is +ve
If we add them we get a nice expression ( ref figure)
It has two parts (i) (1)^(k+1) we see that the answer will be +1 when k is odd and 1 when k is even
Why ? (1) ^even = +1 and (1)^odd = 1 , but when k = even k+1 will be odd and when k = odd k+1 will be even
(ii) so there will +ve values as well as ve value
From (i) we see that for even k the given function is ve and for odd k the given function is +ve
If we add them we get a nice expression ( ref figure)
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when you plug in k=1, you get 1/2. as k increases shouldn't the value of the formula decrease? Hence the answer will never be more than 1/2 so you can eliminate any answer choice that doesn't fit that parameter. Or am I wrong in using that technique?