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

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
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Alternative, if you use the geometric series formula.

S = \frac{a(1-r^n)}{1-r}

where a = first term, r = multiple factor, n = # of terms.

So in this case, a = 1/2, r = -1/2, n = 10.

Therefore,

S=( (1/2) * (1- (-1/2)^10) ) / (1-(-1/2)
S=( (1/2) * (1 - (1/1024) ) / (3/2)
S=(1-1/1024) / 3
S=(1023/1024 ) / 3
Since 1023/1024 is close to 1, dividing it by 3 would get us to approximately 1/3, which is between 1/2 and 1/4. So the answer is D.

Kth term: A(k) = (-1) ^ (k+1) * 1 / 2^k

T is the sum of the first 10 terms of this sequence.

A(1) = (-1)^(1+1)*1/2^1 = 1/2

A(2) = (-1)^(1+2)*1/2^2 = -1/4

You can see that for odd k it would be positive values of 1/2^k and for even k it would be negative values of 1/2^k

So it's like this:

T = \frac{1}{2} - \frac{1}{4} + \frac{1}{18} - \frac{1}{16}+ ...

Now look for a pattern in the sums of various number of terms.

i.e. Sum of first 2 terms = 1/2 - 1/4 = 1/4
Sum of first 3 terms = 1/4 + 1/8 = 3/8
Sum of first 4 terms = 3/8 - 1/16 = 5/16
Sum of first 5 terms = 5/16 + 1/32 = 11/32
and so on...

You can see that all these sums are between 1/4 and 1/2.

On the GMAT when you have tested such a question for about half the terms and you have established a pattern you can safely assume it will continue. This is because this series is analogous to how the GMAT itself adapts the difficulty of its questions based on your answers. It sort of zigzags in a diminishing pattern. So as you increase k, you will be varying only slightly around a sort of ultimate stagnant value at infinity which you could determine as \frac{a}{(1 - r)} i.e. the sum of an infinite Geometric Progression with first term a and common ratio r.

Here we have first term 1/2 and common ratio -1/2

=> Infinite sum = \frac{1}{2*(1 - (-1/2))}= \frac{1}{2*1.5} = \frac{1}{3}

With a large number of terms such as 10 terms, your series sum would actual TEND to stagnate around this infinite sum.

Pick D

When doing sequence problems, it usually helps to look at at least the first few terms. So in this case:

a_k = (-1)^{k+1} * \frac{1}{2^k}

This gives us:

a_1 = \frac{1}{2}
a_2 = -\frac{1}{4}
a_3 = \frac{1}{8}
a_4 = -\frac{1}{16}

We can stop there. The first, and largest, term is 1/2, and we then subtract 1/4. We will then add and subtract fractions that will continue to get smaller and smaller. So we can immediately eliminate A, B, and C, since the sum cannot possibly be greater than 1/2. Now we're left with D and E. Note that the sum of the first two terms is 1/4. Then you add another 1/8 and subtract 1/16. This pattern will continue all the way through the tenth term, and you should be able to see that there's no way this sum will become less than 1/4. So the answer is D.

Hello!

I was just doing this problem and though I would add some "graphic" approach in case it is useful for someone as it has been for me.

You can easily draw a graph of the first points on any series to see if they follow a regular pattern. This one specifically is extremely easy to catch as you draw 3+ points, something like this:

Every odd value of k will make (k+1) even and [(-1)^(k+1)] will be +ve

so we will have

k = 1, [(-1)^(1+1)] × (1 / 2^1) = 1*(1/2) = 1/2
k = 2, [(-1)^(2+1)] × (1 / 2^2) = -1*1/4 = -1/4
k = 3, [(-1)^(3+1)] × (1 / 2^3) = 1 * 1/8 = 1/8

so we can observe we will get alternate +ve and -ve value for (1/2)^k

(1/2) - (1/4) + (1/8) - (1/16) + (1/32) - (1/64) + (1/128) - (1/256) + (1/512) - (1/1024)

[(1/2)+ (1/8) + (1/32)+ (1/128) + (1/512)] - [(1/4)+(1/16)+(1/64)+(1/256)+(1/1024)]

= approx 0.3

so D

The sum of infinite geometric progression with common ratio |r|<1, is sum=\frac{b}{1-r}, where b is the first term.
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HOw to solve in 2 minutes. ?

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

(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

some really good explanations for this question! I was wondering do we need to know the infinite sequence formula - you can solve the question in a minute if you know the formula. The graphic approach was really good too...