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# for every integer k from 1-10 inclusive, the kth term of a

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Senior Manager
Joined: 04 Jun 2007
Posts: 366
for every integer k from 1-10 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 1-10 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 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

Last edited by r019h on 03 Nov 2007, 16:12, edited 2 times in total.
Intern
Joined: 30 Oct 2007
Posts: 9

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31 Oct 2007, 07:27
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(1-q^n)]/(1-q)

T=[1/2*(1-1/2^10)]/(1-1/2) (eliminate the 1/2)
T=1-1/2^10 (fom here we can also eliminate A and B, T must be smaller than 1)
=>T=(2^10-1)/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
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262

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31 Oct 2007, 12:55
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... greater than 1/4
Intern
Joined: 25 Jul 2007
Posts: 16

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01 Nov 2007, 10:10
kevincan wrote:
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... greater than 1/4

How do you get 1/2 - (1/4-1/8) - (1/16-1/32) - ... ? the equation is (-1)^k + 1*(1/2^k)? There should be no negative fractions.
Intern
Joined: 25 Jul 2007
Posts: 16

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01 Nov 2007, 10:25
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
Manager
Joined: 01 Oct 2007
Posts: 87

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01 Nov 2007, 10:38
baileyf16 wrote:
kevincan wrote:
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... greater than 1/4

How do you get 1/2 - (1/4-1/8) - (1/16-1/32) - ... ? 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.
CEO
Joined: 29 Mar 2007
Posts: 2559

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01 Nov 2007, 13:31
johnrb wrote:
baileyf16 wrote:
kevincan wrote:
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... greater than 1/4

How do you get 1/2 - (1/4-1/8) - (1/16-1/32) - ... ? 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.
Intern
Joined: 30 Oct 2007
Posts: 9

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03 Nov 2007, 16:08
johnrb wrote:
baileyf16 wrote:
kevincan wrote:
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... greater than 1/4

How do you get 1/2 - (1/4-1/8) - (1/16-1/32) - ... ? 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+dn-d) will work here.
Intern
Joined: 06 Oct 2007
Posts: 15

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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
Manager
Joined: 01 Nov 2007
Posts: 69

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05 Nov 2007, 10:39
kevincan wrote:
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... greater than 1/4

I think this is the best way this can be explained......thanks
Senior Manager
Joined: 29 Jan 2011
Posts: 358

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24 Jul 2011, 18:25
sportyrizwan wrote:
kevincan wrote:
r019h wrote:
for every integer k from 1-10 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 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

1/2 -(1/4-1/8)-(1/16-1/32)-...less than 1/2

(1/2-1/4) +(1/8-1/16)+... 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   [#permalink] 24 Jul 2011, 18:25
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