GMAT PREP QUESTION w/OA

equation is [(-1)^(k+1)]*(1 / 2^k)

for k = 1

you will get: [(-1)^2]*(1/2) = 1*1/2=1/2

k=2 you will get (-1)*1/4 = -1/4

k=3 you will get (1)* 1/8 = 1/8

k=4 you will get -1/16

k=5 you will get 1/32

and so on......

Note your series would look something like this

1/2 - 1/4 + 1/8 - 1/16 + 1/32..........

basically, you are subtracting 1/4 from 1/2 you get 1/4

then you add 1/8 to 1/4 = 3/8...

all in all, your end result for the series should be between 1/2 and 1/4

rohangupta83 wrote:equation is [(-1)^(k+1)]*(1 / 2^k)

for k = 1

you will get: [(-1)^2]*(1/2) = 1*1/2=1/2

k=2 you will get (-1)*1/4 = -1/4

k=3 you will get (1)* 1/8 = 1/8

k=4 you will get -1/16

k=5 you will get 1/32

and so on......

Note your series would look something like this

1/2 - 1/4 + 1/8 - 1/16 + 1/32..........

basically, you are subtracting 1/4 from 1/2 you get 1/4

then you add 1/8 to 1/4 = 3/8...

all in all, your end result for the series should be between 1/2 and 1/4

I don't even understand what the questions is asking for. The language is very odd. And how would you do your method in under 2 minutes?

well, the question says that:

For any integer k from 1 to 10, inclusive - which means k could be 1-10 (inclusive)

kth of a certain sequence is given by [(-1)^(k+1)]*(1 / 2^k) - which means that kth term is given by the above sequence

For example you want to find out the first term of the sequence (remember, there are only 10 terms in the sequence with k from 1-10) i.e. k=1

plug k = 1 into the above formual and you get the first term and so on

Finally - the question is asking you to find the sum of first 10 terms of this sequence i.e. find the sum of the sequence given by the formula [(-1)^(k+1)]*(1 / 2^k) where 1<= k <=10

Hopefully this should clear what the question is asking you to do.

and in the solution, all you need are the first 4 terms at most to give you an idea what approximately is your answer.

For example

lets assume that the first term is 100

you take half out i.e subtract 50 so now you are left with 50

now you are adding 25 to this 50 - you have 75

now subtracting 12.5 from 75 - you have 62.5

and if you continue the same series which is adding half of what you just subtracted to the result and then subtracting half of what you just added from the result, you will notice that you will oscillate between the range of 50-100, never crossing either extremes.

It is the same principle implied in the above problem

This is an approximation question, hope this helps.

T = 1/2 - 1/2^2 + 1/2^3 - ... -1/2^10

= 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5

Notice that 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 < 1/4, we can say that 1/4 < T < 1/2.

Can someone please clarify how to solve this problem with 2 mins:
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:

logitech wrote:This is an approximation question, hope this helps.

You can use the formula for a geometric progression if you want.

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:

It has two parts (i) (-1)^(k+1) we see that the answer will be +1 when k is odd and -1 when k is even

Why ? (-1) ^even = +1 and (-1)^odd = -1 , but when k = even k+1 will be odd and when k = odd k+1 will be even

(ii) so there will +ve values as well as -ve value

From (i) we see that for even k the given function is -ve and for odd k the given function is +ve

If we add them we get a nice expression ( ref figure)

(1/2 + 1/2^3 +1/2^5....) - 1/2 (1/2 + 1/2^3 + ...)

= 1/2(1/2 + 1/2^3 + ...)
= 1/4 + 1/4( 1/4 + 1/2^4...)
= 1/4 + 1/4( value will bw < 1)

Sum ranges between 1/4 and 1/2

Thanks
Raama

when you plug in k=1, you get 1/2. as k increases shouldn't the value of the formula decrease? Hence the answer will never be more than 1/2 so you can eliminate any answer choice that doesn't fit that parameter. Or am I wrong in using that technique?