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

This is GMATPREP question. What is the best approach to solve this problem quicker?

Hey, couple of things:

- is it supposed to say "the kth NUMBER" or something like that?
- is this bit (1 / 2^k) supposed to read (1/2)^k or 1/(2^k)?

The formatting here is pretty annoying - this one will qualify for a screen shot, if you want to take the time to do that.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

as with most problems asking for the 10th (or 50th, or 408th, or ...) item in some sequence, the key to this one is to look for a pattern in the first few terms, and then extrapolate that pattern.

first term (= 'sum of first 1 terms') = +1/2
sum of first two terms = +1/2 - 1/4 = +1/4
sum of first three terms = +1/2 - 1/4 + 1/8 = +3/8
sum of first four terms = +1/2 - 1/4 + 1/8 - 1/16 = +5/16
sum of first five terms = ... + 1/32 = 11/32
... etc

notice that, from the sum of 3 terms onward, everything you'll get is between 1/4 and 1/2. once you see 3-4 repeats of this result, you can rest assured that the pattern will continue, so the answer is d.

---- nothing below this line is essential ----

theory corner:
you're adding / subtracting smaller and smaller values every time, which means you're moving in a smaller and smaller 'zigzag' around some eventually limiting value. (if you're a student in our course, the idea is markedly similar to the graphs we showed you when we talked about how the cat exam zeroes in on your ability level.) because of the diminishing value of the increments, results #3 and #4 above provide upper and lower bounds on all later values. because both of those values lie between 1/4 and 1/2, there's your answer.

theory corner #2:
if you know the formula for the sum of a geometric series (a/(1 - r)), you can figure out that the sum of the entire series (if it's continued out to infinity) is (1/2)/(1 + 1/2) = 1/3. that value is between 1/4 and 1/2; and, after ten rapidly diminishing terms, you know you're going to be pretty close to that.

Ron or anyone else - Can you explain how this formula works for a geometric series, possibly an example? Thanks!

i will, with the caveat that it will be absolutely unnecessary for any problem appearing on the official gmat.

some geometric series have an infinite number of terms. in fact, whether you realize it or not, you've known this for quite a long time already - because repeating decimals are nothing other than infinite geometric series in disguise. for instance,
0.33333... = 3/10 + 3/100 + 3/1000 + ...
which is a geometric series with a = 3/10, r = 1/10, and an infinite # of terms.

the formula for the sum of an infinite geometric series is a/(1 - r), where a is the first term and r is the ratio of consecutive terms. so, if we apply this formula to the series given above, we get
(3/10) / (1 - 1/10) = (3/10) / (9/10)
= 3/9
= 1/3, as we knew all along.

why does the first term have to be 1/2, can it be -1/2 is k is an odd number?

i mean even number.

the FIRST term means that k = 1, which is definitively an odd number.

a^n+a^(n-1)+...a+1=(a^(n+1)-1)/(a-1)
example: 1+3^1+3^2+...3^9=(3^10-1)/(3-1)

T=1/2-1/2^2+1/2^3-...+1/2^10
T=1/2*(1+(-1/2)+(-1/2)^2+...+(-1/2)^9)


T=1/2*(1-(-1/2)^10)/(1-(-1/2))
T=(2^10-1)/(3*2^10)
2^10 >>1 ---> T=/=1/3=0.3333

Thanks elgharniti. I believe you provided a rephrase of the formal formula for the geometric series:

Sn = (a)[(1 - r^n)/(1 - r)]

Where Sn is the sum of n terms in a geometric sequence, a is the first term of the series, r is the common ratio, and n is the number of terms in a geometric sequence you want to add up.

(The formula Ron provided, S = a / (1 - r) is useful for an INFINITE geometric series where | r | < 1. It doesn't work when r > 1 because the series diverges, or doesn't do the "zig zag" pattern; it just keeps getting bigger).

I don't know whether this formula helps on the actual GMAT, but you will never need to use it to solve any question on the GMAT.

_________________
Ben Ku
Instructor
ManhattanGMAT