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for every integer k from 1-10 inclusive, the kth term of a [#permalink]
30 Oct 2007, 19:14

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (01:02) correct
0% (00:00) wrong based on 0 sessions

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

please explain

Last edited by r019h on 03 Nov 2007, 15:12, edited 2 times in total.

Re: GMATPREP SEQUENCE PS [#permalink]
31 Oct 2007, 11:55

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

Re: GMATPREP SEQUENCE PS [#permalink]
01 Nov 2007, 09:10

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

please explain

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.

Re: GMATPREP SEQUENCE PS [#permalink]
01 Nov 2007, 09:38

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

please explain

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:

Re: GMATPREP SEQUENCE PS [#permalink]
01 Nov 2007, 12:31

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

please explain

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.

Re: GMATPREP SEQUENCE PS [#permalink]
03 Nov 2007, 15:08

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

please explain

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.

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!

Re: GMATPREP SEQUENCE PS [#permalink]
05 Nov 2007, 09:39

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

please explain

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

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

please explain

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

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