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 Post subject: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Sun Apr 12, 2009 1:49 pm 
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Course Students


Posts: 1
certain sequence is given by (-1)^k+1 (1/2^k). If T is the sum of the first 10 terms in 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

Answer D


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 Post subject: Re: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Sun Apr 12, 2009 3:23 pm 
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Posts: 18
Quote:
For every integer k from 1 to 10, the kth term of a certain sequence is given by (-1)^k+1 (1/2^k). If T is the sum of the first 10 terms in 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

Answer D


(-1)^k+1 (1/2^k)

1st term K=1 : -1^2(1/2^1)=1/2
2nd term K=2 : -1^3(1/2^2)=-1/4
so when k=odd sign is positive k=even sign is negative; and it multiply
so the sequence is : 1/2. -1/4, 1/8, -1/16, 1/32, -1/64, 1/128, -1/256, 1/512, -1024.

now add all those you will get the desire result.

if anyone can show some quick trick it will be helpful

thanks
MD.SHOFIKUL ISLAM


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 Post subject: Re: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Tue Apr 14, 2009 6:25 pm 
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Forum Guests


Posts: 6
Your expression of (-1)^k+1 (1/2^k) is very confusing, wrong as a matter of fact. Put this way, the expression equals 0 for any odd k and equals 2 for any even k. Try using something like "x^y, where x = ..., and y = ..." to make it clearer.

And then we can try to tackle the question.


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 Post subject: Re: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Wed Apr 15, 2009 11:14 pm 
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ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
I see what Joey is saying. However, the addition of some parentheses would clarify sufficiently. The sequence definition is [(-1)^(k+1)] * (1/2^k).

I would basically approach this one as Shofikul did, computing at least the first 4 or 5 terms.
shobujgmat wrote:
1st term K=1 : -1^2(1/2^1)=1/2
2nd term K=2 : -1^3(1/2^2)=-1/4
so when k=odd sign is positive k=even sign is negative; and it multiply
so the sequence is : 1/2. -1/4, 1/8, -1/16, 1/32, -1/64, 1/128, -1/256, 1/512, -1024.

now add all those you will get the desire result.

if anyone can show some quick trick it will be helpful

Quick tricks, as requested:
(1) You really don't need to write all ten terms, as you will see the pattern emerge (alternating signs +/-, fractions decreasing in absolute value)

(2) Pair terms as you sum and you'll see patterns there, too.

One way: (1/2-1/4) + (1/8-1/16) + (1/32-1/64) + etc. = 1/4 + 1/16 + 1/64 + etc.
----> sum of only positive fractions, so the sum > 1/4.

Another way: 1/2 + (-1/4+1/8) + (-1/16+1/32) + etc. = 1/2 - 1/8 - 1/16 - etc.
----> 1/2 minus some fractions, so the sum < 1/2.

In this way, we can determine a max and min for the exact sum.

(3) A visual approach: On a number line, move your pencil as you sum through the list. Go to 1/2, then down to 1/4, then up to 3/8, then down to 5/16, etc. You'll see that your pencil moves less each time, and is converging on some number between 1/4 and 1/2.

_________________
Emily Sledge
Instructor
ManhattanGMAT


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 Post subject: Re: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Sun Aug 15, 2010 4:32 pm 
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Course Students


Posts: 1
Can you use the formula - avg times the number of terms, and get the average by the first and last terms divided by 2??


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 Post subject: Re: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Tue Aug 24, 2010 5:11 pm 
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Students


Posts: 2
Emily,

I applied your approach exactly.

But I still thought it was too time consuming while I was taking the test.

Is there a faster approach ?

Thank you,

Anwesha Deb.


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 Post subject: Re: PS: For every integer k from 1 to 10, the kth term of a
 Post Posted: Wed Sep 01, 2010 5:16 am 
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ManhattanGMAT Staff


Posts: 7146
anwesha.deb wrote:
Emily,

I applied your approach exactly.

But I still thought it was too time consuming while I was taking the test.

Is there a faster approach ?


there is, if you know a certain formula -- this type of sum is called a "geometric series", so, if you know the formula for a geometric series, you can find the sum.

in this sort of series, each term is the same multiple "r" of the previous term. note that every term of this series is -1/2 times the previous term:
1/2
-1/4
1/8
-1/16
etc.
so "r" = -1/2.
also, "a", the first term, is 1/2 as well.

the formula says that, if there are "n" terms in the sequence, then the sum of all of them is
a*(1 - r^n)/(1 - r)
= (1/2)*(1 - (-1/2)^10)/(1 - (-1/2))
= (1/2)(1023/1024)/(3/2)
= 1023/3072
this is between 1/4 and 1/2.

of course, that's probably not any better than just straight-up adding the terms together, but ... hey, you asked for it.

--

in any case, it's very likely (close to 0% probability) that you're going to see any other problems that are purely the sum of a geometric series, but it's highly likely that you'll see something like this (i.e., some sort of multi-step pattern recognition problem) again.

takeaway:
if any problem has TOO MANY STEPS TO BE REASONABLE, or if the behavior is TOO WEIRD for you to figure out the appropriate theory, then you should be able to do PATTERN RECOGNITION.


on a gmat problem, you will never HAVE TO do the same sort of step more than 2-3 times -- this is not a "busy work" sort of test. (this doesn't mean that you can't do more than 2-3 of the same step -- it just means that there will always be some type of approach that doesn't.)

in other words:
START WORKING TOWARD THAT GOAL -- even though you know that the goal itself is WAY too far away to reach -- until you SEE A PATTERN.

ONCE YOU SEE A PATTERN, JUST CONTINUE THE PATTERN. DO NOT WORRY ABOUT WHY THE PATTERN EXISTS.


in this problem:
if you add up the first 3 terms, you'll get a value between 1/2 and 1/4.
if you add up the first 4 terms, you'll get a value between 1/2 and 1/4.
if you add up the first 5 terms, you'll get a value between 1/2 and 1/4.
etc.
you can trust that this pattern will continue.


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