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sequence

by paresh_patil » Sat Mar 16, 2013 12:34 am
For every integer k from 1 to 10, inclusive, the k th 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 & 2
C) between 1/2 & 1
D) between 1/4 & 1/2
E) less than 1/4
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by Anju@Gurome » Sat Mar 16, 2013 12:49 am
paresh_patil wrote:For every integer k from 1 to 10, inclusive, the k th 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
k th term = (-1)^(k+1)*((1/2)^k)

Hence,
  • 1st term = (-1)^(1+1)*((1/2)^1) = 1/2
    2nd term = (-1)^(2+1)*((1/2)^2) = -1/4
    3rd term = (-1)^(3+1)*((1/2)^3) = 1/8
    4th term = (-1)^(4+1)*((1/2)^4) = -1/16
    Etc ...
Now, the sum of first 10 terms is nothing but repeated subtraction and addition (in that order) of half of the previous term.

To understand this more try to visualize the situation on the number line. Start from 1/2 and subtract 14 from it. So we are at 1/4 now. Add 1/8 to it. We will be definitely somewhere less than 1/2 now (1/4 + 1/8 = 3/8). Subtract 1/16 from it. We will be definitely somewhere greater than 1/4 now (3/8 - 1/16 = 5/16). Now the sum is nothing but repetition of this pattern.

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Hence, T will be definitely less than 1/2 and greater than (1/2 - 1/4) = 1/4

The correct answer is D.
Last edited by Anju@Gurome on Sat Mar 16, 2013 1:08 am, edited 1 time in total.
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by Anju@Gurome » Sat Mar 16, 2013 12:56 am
Algebraic Approach #1:
The sum of n consecutive 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, so T = 1/3, which lies between 1/4 and 1/2.

The correct answer is D.
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by Anju@Gurome » Sat Mar 16, 2013 1:16 am
Algebraic Approach #2:
k-th term = ((-1)^(k+1)*((1/2)^k)
Hence,
  • 1st term = (-1)^(1+1)*((1/2)^1) = 1/2
    2nd term = (-1)^(2+1)*((1/2)^2) = -1/4 = -(1/2)^2
    3rd term = (-1)^(3+1)*((1/2)^3) = 1/8 = (1/2)^3
    etc...
So, T = 1/2 - (1/2)^2 + (1/2)^3 - (1/2)^4 + ... up to 10 terms
--> T = (1/2 + 1/2^3 + 1/2^5 + ... ) - (1/2^2 + 1/2^4 +... )
--> T = (1/2)*(1 + 1/2^2 + 1/2^4 + ... ) - (1/2^2)*(1 + 1/2^2 + 1/2^4 +... )
--> T = (1/2 - 1/4)*(1 + 1/2^2 + 1/2^4 + ... )
--> T = (1/4)*(1 + 1/2^2 + 1/2^4 + ... )

Now, 1 < (1 + 1/2^2 + 1/2^4 + ... ) < 2

Hence, 1/4 < T < 1/2

The correct answer is D.
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by GMATGuruNY » Sat Mar 16, 2013 6:02 am
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by Brent@GMATPrepNow » Sat Mar 16, 2013 6:08 am
paresh_patil wrote:For every integer k from 1 to 10, inclusive, the k th 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 & 2
C) between 1/2 & 1
D) between 1/4 & 1/2
E) less than 1/4
As we've already seen, T = 1/2 - 1/4 + 1/8 - 1/16 + . . .

Here's another way to evaluate this mess.

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

Then, when we start simplifying each part in brackets, we'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)
So, T = 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
Brent Hanneson - Creator of GMATPrepNow.com
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