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For every integer k from 1 to 10...

by NYC493 » Mon Mar 05, 2012 2:22 pm

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Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
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by pemdas » Mon Mar 05, 2012 2:42 pm
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
you can find many interesting solutions on BTG
i'll offer mine and will try to renew this question's thread :)

you are right the sign alternates and we get something like 1/2 -1/4 +1/8 -1/16 ... or (2^9 -2^8 +2^7 -2^6 +2^5 -2^4 ...)/2^10 by further simplification we get [2^8(2-1) +2^6(2-1) +2^4(2-1) ...)/2^10= (2^8 + 2^6 + 2^4 + 2^2)/2^10 = 2^2(2^6 + 2^4 +2^2 +1)/2^10 = (2^6 + 2^4 +2^2 +1)/2^8 or (64+16+4+1)/(16*16) [spoiler]2^8=(2^4)*(2^4)=16*16[/spoiler], 85/256 =~1/3

d
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by krusta80 » Mon Mar 05, 2012 3:25 pm
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Pemdas's method is very impressive...a lot can be learned from the way he manipulated the algebra. That said, though, during a test -- when time is of the essence -- I would probably do something along the following lines...


STEP 1

Formulas of these types can be generalized as...

(-1)^(i) * (r)^j

The first thing I look at is the value of r. Specifically, is r less than 1 or greater than 1? If it's less than 1 (btw...it almost always will be in my experience), then I know that the sum will change less and less as j increases. This is because a fraction less than 1 will get smaller and smaller as it's exponent increases.

The next thing to do is to figure out how many terms are enough to determine the right choice. I do this by first finding the smallest range from the given choices. In this case...

A) Range is infinite
B) Range is 1
C) Range is 1/2
D) Range is 1/4
E) Range is 1/2

***Side note: The GMAT is designed to challenge us...to trick us at times. Therefore, right away I'm leaning toward either C,D, or E. Why? Because their target ranges are the smallest, and the smaller the range, the more terms required to determine whether that answer choice is correct.

So, the smallest range is 1/4. What power of 1/2 gives us 1/4? 2 of course. This means that it'll take two terms to reach that range's level of precision. For this problem:

1/2 - 1/4 = 1/4 OK, so we're close already. It's either going to be D or E. :)

Once you reach the required level of precision, I would be safe by going with two more terms (notice always an EVEN number of terms to cancel out the opposing signs):

1/4 + 1/8 - 1/16 = (4+2-1)/16 = 5/16.

This leaves choice D. Hey what do you know...they picked the narrowest range!
Last edited by krusta80 on Mon Mar 05, 2012 6:37 pm, edited 1 time in total.
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by NYC493 » Mon Mar 05, 2012 4:37 pm
Thank you both! I'll have to keep practicing problems of this type. But, your strategies are appreciated!
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by Anurag@Gurome » Mon Mar 05, 2012 7:05 pm
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
We know that the sum of n terms of a geometric series is given by:
S(n) = a(1 - r�)(1 - r), where a is the first term, r is the common ratio of the geometric progression and n = number of terms.

Here, a = 1/2, r = -1/2, n = 10
T = 1/2[1 - (-1/2)^10]/[1 + 1/2]
= 1/2[1 - 1/1024]/[3/2]
= 1/2 * 1023/1024 * (2/3)
= (1023/1024) * (1/3)
Now (1023/1024) = 1 approx, so T = 1/3, which lies between 1/4 and 1/2.

The correct answer is D.
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by aircraft » Mon May 20, 2013 3:34 pm
Excuse me...could it be possible that Mr Mairal means S(n) = a(1 - r�)/(1 - r) instead of S(n) = a(1 - r�)(1 - r).

Thanks!
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by Brent@GMATPrepNow » Mon May 20, 2013 9:58 pm
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
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by gmattesttaker2 » Fri Jul 26, 2013 6:21 pm
Brent@GMATPrepNow wrote:
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
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by gmattesttaker2 » Fri Jul 26, 2013 6:22 pm
Brent@GMATPrepNow wrote:
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
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by gmattesttaker2 » Fri Jul 26, 2013 6:23 pm
Brent@GMATPrepNow wrote:
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
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by gmattesttaker2 » Fri Jul 26, 2013 6:53 pm
Brent@GMATPrepNow wrote:
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
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by gmattesttaker2 » Fri Jul 26, 2013 6:58 pm
Brent@GMATPrepNow wrote:
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
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by gmattesttaker2 » Fri Jul 26, 2013 7:00 pm
Brent@GMATPrepNow wrote:
NYC493 wrote:Can someone tell me a quick way to decode a problem like this? If you start to solve each number in the sequence manually, you can see that it's alternating from positive to negative. But, how do you arrive at an approximate sum quickly?

"For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (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 1/2

Thanks!
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
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by Brent@GMATPrepNow » Sat Jul 27, 2013 6:08 am
gmattesttaker2 wrote:I was clear till this point:
Of course, we can also see that T > 1/4
Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri
Hi Sri,

Once we get to this point... T = 1/4 + (a number less 1/4) ..., we can say that T = 1/4 + (some positive value).
If we add some positive value to 1/4, the sum (T) must be greater than 1/4

I hope that helps.

Cheers,
Brent
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by hemant_rajput » Sun Jul 28, 2013 8:54 am
here is my approach.

so first list down the term say 5 terms and see the pattern.

(1/2) , -(1/4),(1/8), -(1/16), (1/32)

so sum will be 1/2 - 1/4 + 1/8 - 1/16 +1/32...

add the number in pair of 2
so new series will be [1/2 - 1/4] + [1/8 - 1/16] +[ 1/32 - 1/64]...

1/4 + 1/16 + 1/64 ...
this is an infinite series and a Geometric progression too.

you can always use this simple formula to find the sum of a infinite GP series.

a/(1 - r), where a is the first term and r is the progressing factor.

a = 1/4

r = (1/4) / (1/16) = 1/4

so answer will be (1/4)/ [1 - (1/4)] = (1/4)/(3/4) = 1/3

which satisfy option D
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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