sum of terms in a sequence

[josh80]

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1)(1/2^k). If T is the sum of first 10 terms in the sequence, then T is

greater than 2
between 1 and 2
between 1/2 and 1
between 1/4 and 1/2
less than 1/4

[Brent@GMATPrepNow]

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1)[(1/2)^k)]. If T is the sum of 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

One way to solve this question is to apply the formula for what's known as a "geometric series," but I'm not really a fan of memorizing formulas. Another option is to get a better idea of this sum, by first 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 than 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

[Amrabdelnaby]

Im little bit confused regarding this question.

could you please explain how did you arrive to the first few terms?

also this is a negative number. i thought the outcome will be once negative and once positive.

the first term for example would be -1^(k+1)(1/2)^k = -1(2)(1/2) = -1.. so how come the first term is 1/4?

please explain

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1)[(1/2)^k)]. If T is the sum of 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

One way to solve this question is to apply the formula for what's known as a "geometric series," but I'm not really a fan of memorizing formulas. Another option is to get a better idea of this sum, by first 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 than 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

[GMATGuruNY]

For every integer K from 1 to 10, inclusive the kth term of a certain sequence is given by (-1)^(k+1) (1/2^K).
If T is the sum of first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Calculate until you see the pattern.
Some test-takers might find it helpful to visualize the sum on a number line.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (1/16).
In other words, the sum will alternate between increasing a little and then decreasing a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Here are the first four terms, plotted on a number line:



Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

[Amrabdelnaby]

Ok I think I understand why i got it wrong now. thought that (k+1)(1/2^k) are both exponents of -1 while the fact is: -1^(k+1) x (1/2^k)

hence for the first term it is -1^2 x (1/2) which is basically 1 x 1/2 = 1/2

Thanks Guru

For every integer K from 1 to 10, inclusive the kth term of a certain sequence is given by (-1)^(k+1) (1/2^K).
If T is the sum of first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Calculate until you see the pattern.
Some test-takers might find it helpful to visualize the sum on a number line.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (1/16).
In other words, the sum will alternate between increasing a little and then decreasing a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Here are the first four terms, plotted on a number line:



Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

[Brent@GMATPrepNow]

Im little bit confused regarding this question.

could you please explain how did you arrive to the first few terms?

also this is a negative number. i thought the outcome will be once negative and once positive.

the first term for example would be -1^(k+1)(1/2)^k = -1(2)(1/2) = -1.. so how come the first term is 1/4?

please explain

kth term of a certain sequence = (-1)^(k+1) (1/2^k)

term1 = (-1)^(1+1) (1/2^1)
= [(-1)^2](1/2)
= [1](1/2)
= 1/2

term2 = (-1)^(2+1) (1/2^2)
= [(-1)^3](1/4)
= [-1](1/4)
= -1/4


term3 = (-1)^(3+1) (1/2^3)
= [(-1)^4](1/8)
= [1](1/8)
= 1/8

Etc.

So, we can write: T = 1/2 + (-1/4) + 1/8 + (-1/16) ....
Or we can write: T = 1/2 - 1/4 + 1/8 - 1/16 ....

Cheers,
Brent

[[email protected]]

Hi All,

This question comes up every so often in this Forum. There are a couple of different ways of thinking about this problem, but they all require a certain degree of "math."

You've probably correctly deduced what the sequence is:

+1/2, -1/4, +1/8, -1/16, etc.

The "key" to solving this question quickly is to think about the terms in "sets of 2"...

1/2 - 1/4 = 1/4

Since the first term in each "set of 2" is greater than the second (negative) term, we now know that each set of 2 will be positive.

1/8 - 1/16 = 1/16

Now we know that each additional set of 2 will be significantly smaller than the prior set of 2.

Without doing all of the calculations, we know....
We have 1/4 and we'll be adding tinier and tinier fractions to it. Since there are only 10 terms in the sequence, there are only 5 sets of 2, so we won't be adding much to 1/4. Based on the answer choices, only one answer makes any sense...

Final Answer: D

GMAT assassins aren't born, they're made,
Rich