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then T is... (GMAT Prep)

by alex.gellatly » Mon Jul 02, 2012 9:33 pm
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 the 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
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by tutorphd » Mon Jul 02, 2012 9:48 pm
It is not clear whether you mean (-1)^k + 1/2^k or (-1)^(k+1) * (1/2^k)?

The solution below assumes the former.
The sign changing terms (-1)^k are going to cancel out because you will have 5 that are +1 and 5 that are -1.

You are left with T = 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + ... + 1/2^10

1/2 < T because the first term is 1/2 and you are adding positive numbers on top of it.

Draw the number axis and add graphically length segments corresponding to the different fractions in T = 1/2 + 1/4 + 1/8 + 1/16 + .... , the 1/2 segment starting from 0, each subsequent segment starting where the previous ends - the full length covered is a graphical visualization of the running total. Each subsequent fraction adds haft of the remaining segment to 1. You will see visually that the sum is approaching 1 from below when you increase the number of terms.

so 1/2 < T < 1
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by Anurag@Gurome » Mon Jul 02, 2012 10:38 pm
k-th term = [(-1)^(k+1)]*[(1/2)^k]
1st term = [(-1)^(1+1)]*[(1/2)^1] = 1/2
2nd term = [(-1)^(1+2)]*[(1/2)^2] = -(1/2)^2

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 + (1/2)^2 + (1/2)^4 + ..] is greater than 1 but less than 2.
Therefore, 1/4 < T < 1/2

The correct answer is D.
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by Anurag@Gurome » Mon Jul 02, 2012 10:43 pm
This problem can also be solved using the formula of geometric progression.
The sum of n terms of a geometric series is given by:
  • S(n) = a(1 - râ�¿)(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
Hence, T = 1/3, which lies between 1/4 and 1/2.

The correct answer is D.
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by GMATGuruNY » Tue Jul 03, 2012 1:51 am
alex.gellatly wrote: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 the 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
Notice that the answer choices are RANGES.
We are not expected to calculate the exact sum.
Use a NUMBER LINE to determine the correct range.

Image

Follow the arrows.
The first term is 1/2.
When we add in -1/4 -- the second term -- the sum decreases to 1/4.
When we add in +1/8 -- the third term -- the sum increases to 3/8.
By now, we can already see that the sum will converge to a value somewhere between 1/4 and 3/8.

The correct answer is D.
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