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

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Intern
Joined: 03 May 2008
Posts: 6
For every integer k from 1 to 10, inclusive, the kth term of [#permalink]

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31 Aug 2008, 01:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For every integer k from 1 to 10, inclusive, the kth term of a sequence is given by (-1)raised to k+1 (1/2 raised to 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 ½ and 1
D. between ¼ and ½
E. less than ¼

Current Student
Joined: 11 May 2008
Posts: 556

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31 Aug 2008, 02:26
hmmm... D???
is it correct?
Intern
Joined: 03 May 2008
Posts: 6

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31 Aug 2008, 02:51
yep. can u explain the steps pls
SVP
Joined: 07 Nov 2007
Posts: 1801
Location: New York

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31 Aug 2008, 08:05
cyclops wrote:
For every integer k from 1 to 10, inclusive, the kth term of a sequence is given by (-1)raised to k+1 (1/2 raised to 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 ½ and 1
D. between ¼ and ½
E. less than ¼

expand thes series

= 1/2 -1/2^2 +1/2^3-1/2^4.... -1/2^10
= [1/2+1/2^3+1/2^5+1/2^7+1/2^9]-1/2[1/2+1/2^3+1/2^5+1/2^7+1/2^9]
=1/2*[1/2+1/2^3+1/2^5+1/2^7+1/2^9]
= 1/2* [Value between 1/2 and 1]
= Value between 1/4 and 1/2
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Last edited by x2suresh on 31 Aug 2008, 09:33, edited 2 times in total.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1179

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31 Aug 2008, 08:35
cyclops wrote:
For every integer k from 1 to 10, inclusive, the kth term of a sequence is given by (-1)raised to k+1 (1/2 raised to 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 ½ and 1
D. between ¼ and ½
E. less than ¼

You could learn a formula for such questions (this is an alternating geometric series), but you don't need to. It's almost always a good idea to write out at least the first few terms of a sequence- often you'll notice a pattern that will help to answer the question. If you do this, you'll see that the question is asking:

1/2 - 1/4 + 1/8 -1/16 + ... + 1/(2^9) - 1/2^(10) = ?

There are many ways to estimate the value of this sum. You can notice, for example, that 1/2 - 1/4 = 1/4; 1/8 - 1/16 = 1/16, and so on, so the sum is exactly equal to:

1/4 + 1/16 + 1/64 + 1/256 + 1/1028

and which is clearly only slightly greater than 1/4.

There's another way you could look at this, though it becomes much more clear when you can draw everything on a number line, which I can't do here. Label the points 0 and 1 on a number line. Suppose you're going to do the following:

-run halfway from 0 to 1, arriving at A: then A is 1/2
-run backwards half the distance between A and 0, arriving at B: then B is 1/2 - 1/4
-run forwards half the distance between B and A, arriving at C: then C is 1/2 - 1/4 + 1/8
-run backwards half the distance between C and B, arriving at D: then D is 1/2 - 1/4 + 1/8 - 1/16
-etc.

you'll see that no matter how many terms you add of this sequence, the sum must be between 1/4 and 1/2.
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SVP
Joined: 21 Jul 2006
Posts: 1510

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31 Aug 2008, 09:21
Actually, if you try to draw the number line, this can be really easy. first of all, from the sequence formula, our sequence will be:

1/2, -1/4, 1/8, -1/32, .....etc, so as you can see, the number is just getting smaller and smaller. so when you draw the number line, you can draw it like this:

0----1/4----1/2----3/4----1

so start from 1/2, when you subtract 1/4, you'll land on 1/4, then when you add 1/8, which is smaller than 1/4 and 1/2, you will land somewhere before 1/2, then when you subtract 1/32, which is even smaller than 1/8, 1/2, and 1/4, you will land somewhere behind but still be somewhere between 1/4 and 1/2. So eventually, you're just jumping back and forth in the area between 1/4 and 1/2 because the new fraction that you add is just getting smaller and smaller.

do you guys see what I mean? so that's how we end up with D as our answer.

hope this can help
Current Student
Joined: 11 May 2008
Posts: 556

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31 Aug 2008, 09:36
ian ... a diff approach... thats what we need here +1
Intern
Joined: 03 May 2008
Posts: 6

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04 Sep 2008, 08:10
Thanks guys for the explanation.
Re: gmatprep math   [#permalink] 04 Sep 2008, 08:10
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