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for every integer k from 110 inclusive, the kth term of a [#permalink]
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30 Oct 2007, 20:14
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for every integer k from 110 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 ten terms in the sequence, then T is
greater than 2
between 1 & 2
between 1/2 and 1
between 1/4 and 1/2 < answer (but do we have to write out all the terms and then sum them up? that would be tedious)
less than 1/4
please explain
Last edited by r019h on 03 Nov 2007, 16:12, edited 2 times in total.



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Joined: 30 Oct 2007
Posts: 9

C between 1/2 and 1
if i am right the question can be sloved as following
for the question already gives u An=(1)^k+1*(1/2^k).
=>A1=(1)^1+1*(1/2^1) => A1=1+1/2
A2=1+1/2^2
A3=1+1/2^3
.......
A9=1+1/2^9
A10=1+1/2^10
u can get 5 positive 1 and 5 negative 1(1)
So the T=A1+A2+A3....+A10=1+1/2+1+1/2^2+.......+(1)+1/2^9+1+1/2^10
eliminate the 1 and 1
u get T=1/2+1/2^2+1/2^3+.......+1/2^9+1/2^10 (This is a geometric progression/sequence)
the equation can be set up T=[A1(1q^n)]/(1q)
T=[1/2*(11/2^10)]/(11/2) (eliminate the 1/2)
T=11/2^10 (fom here we can also eliminate A and B, T must be smaller than 1)
=>T=(2^101)/2^10
for 1/4 we can change it into 1/2^2=>2^8/2^2*2^8=>2^8/2^10
apparently T is larger than 1/4
for 1/2 also change it into 2^9/2^10 also smaller than T
So 1/2<T<1



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Re: GMATPREP SEQUENCE PS [#permalink]
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31 Oct 2007, 12:55
r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain
1/2 (1/41/8)(1/161/32)...less than 1/2
(1/21/4) +(1/81/16)+... greater than 1/4



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Joined: 25 Jul 2007
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Re: GMATPREP SEQUENCE PS [#permalink]
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01 Nov 2007, 10:10
kevincan wrote: r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain 1/2 (1/41/8)(1/161/32)...less than 1/2 (1/21/4) +(1/81/16)+... greater than 1/4
How do you get 1/2  (1/41/8)  (1/161/32)  ... ? the equation is (1)^k + 1*(1/2^k)? There should be no negative fractions.



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Joined: 25 Jul 2007
Posts: 16

1/2 + 1/2^2 + ... + 1/2^10 < 1 since:
1/2 < 1
1/2^2 < 1/2
1/2 + 1/2^2 < 1
1/2^2 + 1/2^3 < 1/2
1/2 + 1/2^2 + 1/2^3 < 1
1/2^2 + 1/2^3 + 1/2^4 < 1/2
1/2 + 1/2^2 +1/2^3 +1/2^4 < 1
.
.
1/2 + 1/2^2 +...+1/2^8 <1
1/2^3 + 1/2^4 + ...+ 1/2^10 < 1/2^2 = 1/4
1/2 + 1/2^2 + ... + 1/2^10 <1.
The series is obviously greater than 1/2 so the answer is C. The exact answer is .999023437



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Re: GMATPREP SEQUENCE PS [#permalink]
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01 Nov 2007, 10:38
baileyf16 wrote: kevincan wrote: r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain 1/2 (1/41/8)(1/161/32)...less than 1/2 (1/21/4) +(1/81/16)+... greater than 1/4 How do you get 1/2  (1/41/8)  (1/161/32)  ... ? the equation is (1)^k + 1*(1/2^k)? There should be no negative fractions.
As written, it's ambiguous, but I think the equation should be bracketed:
[(1)^(k+1)]*[1/(2^k)]. If this is the case, then the sequence is indeed:
1/2  1/4 + 1/8, etc.



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Re: GMATPREP SEQUENCE PS [#permalink]
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01 Nov 2007, 13:31
johnrb wrote: baileyf16 wrote: kevincan wrote: r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain 1/2 (1/41/8)(1/161/32)...less than 1/2 (1/21/4) +(1/81/16)+... greater than 1/4 How do you get 1/2  (1/41/8)  (1/161/32)  ... ? the equation is (1)^k + 1*(1/2^k)? There should be no negative fractions. As written, it's ambiguous, but I think the equation should be bracketed: [(1)^(k+1)]*[1/(2^k)]. If this is the case, then the sequence is indeed: 1/2  1/4 + 1/8, etc.
Yes thank you. I was getting C w/ this equation and I was wondering why the 1 in 1*(1/2^) was even written.



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Joined: 30 Oct 2007
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Re: GMATPREP SEQUENCE PS [#permalink]
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03 Nov 2007, 16:08
johnrb wrote: baileyf16 wrote: kevincan wrote: r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain 1/2 (1/41/8)(1/161/32)...less than 1/2 (1/21/4) +(1/81/16)+... greater than 1/4 How do you get 1/2  (1/41/8)  (1/161/32)  ... ? the equation is (1)^k + 1*(1/2^k)? There should be no negative fractions. As written, it's ambiguous, but I think the equation should be bracketed: [(1)^(k+1)]*[1/(2^k)]. If this is the case, then the sequence is indeed: 1/2  1/4 + 1/8, etc.
how are we getting this? just plugging in values for k?
in which case,
k= 1, we get 1/2
k=2, we get 1/4
k=3, we get 1/8
k=4, we get 1/16
and so on till k= 10, where we get 1/2^10
but from this how do we get the sum of the first ten terms without actually adding everything up? is there a formula we should be using? don't think the AP formula S= n/2(2a+dnd) will work here.



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Joined: 06 Oct 2007
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Same Question... [#permalink]
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05 Nov 2007, 10:16
I took the GMAT prep and had the same question. Is there a specific formula to use for this in order to find the answer?
I think it would be a long procress to add each item esp when it was Question 3 on my test!
Thanks for the help!
~ Ev



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Re: GMATPREP SEQUENCE PS [#permalink]
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05 Nov 2007, 10:39
kevincan wrote: r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain 1/2 (1/41/8)(1/161/32)...less than 1/2 (1/21/4) +(1/81/16)+... greater than 1/4
I think this is the best way this can be explained......thanks



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Re: GMATPREP SEQUENCE PS [#permalink]
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24 Jul 2011, 18:25
sportyrizwan wrote: kevincan wrote: r019h wrote: for every integer k from 110 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 ten terms in the sequence, then T is greater than 2 between 1 & 2 between 1/2 and 1 between 1/4 and 1/2 less than 1/4 please explain 1/2 (1/41/8)(1/161/32)...less than 1/2 (1/21/4) +(1/81/16)+... greater than 1/4 I think this is the best way this can be explained......thanks i DONT UNDERSTAND THIS SOLUTION... Is there a better approach ?? Please help me understand ...in detail




Re: GMATPREP SEQUENCE PS
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24 Jul 2011, 18:25








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