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Exponents

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Exponents

by adi » Sat Mar 08, 2008 2:16 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 items 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 should be D ***

I was able to get that answer but a lot of it was down to guessing and estimating. If someone knows how to get down to the answer properly, please share.
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by rey.fernandez » Sun Mar 09, 2008 12:30 am
Here's one way to go about it:

Let's start by listing out each term in this sequence

k -1^(k+1) * 1/(2^K)
= =============
1 1/2
2 -1/4
3 1/8
4 -1/16
5 1/32
6 -1/64
7 1/128
8 -1/256
9 1/512
10 -1/1024

Now, in truth, you don't need to list out all of these terms and add them up to get to your answer. You can reason your way there by looking at the first few terms.

-you start with 1/2
- subtract 1/4 --> now the value is 1/4
- add 1/8 --> we could calculate it, but qualitatively, we know that adding 1/8 is not enough to get the running total to exceed 1/2. So we know that the current total is above 1/4 and below 1/2.
- subtract 1/16 --> well, now we're subtracting a fairly small amount, not enough to get the running total to go below 1/4.
- add 1/32 --> now we're adding a smaller amount, so there's really no chance that the running total will exceed 1/2.
and so on...

If you carry out this thought experiment for more terms, you see that as you go down the list of terms, you're alternately adding or subtracting successively smaller and smaller amounts and that you remain between the values 1/4 and 1/2 throughout. (These boundaries were created by the first two terms of the series.)
Rey Fernandez
Instructor
Manhattan GMAT
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