Gmat prep problem- 1

sugmomo wrote:hi guys,

i came across this PS while taking gmat prep 1

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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

We don't need to determine the exact sum. Compute only as much as is necessary to see the pattern.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (-1/16, for example).
In other words, the sum will alternate between going up a little and then down a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

We will have a series

1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/ 64 ........................... - 1/1024

we can write it like this


1 - 1/2 + 1/2 - 3/4 + 3/4 - 5/8 + 5/8 - 11/ 16 .......... Last two terms will be 341 / 512 - 681 / 1024


Every thing will cancel out
in the end we will have 1 - 681/1024

it will be 343/1024

and it is indeed between 1/4 and 1/2 .. ( Don't try this at home )
Guru approach is far much more better...

You are great guru........

Thanks Guru


Thanks goyalsau. Can you please elloborate how you arrived to the following step

1 - 1/2 + 1/2 - 3/4 + 3/4 - 5/8 + 5/8 - 11/ 16 .......... Last two terms will be 341 / 512 - 681 / 1024


Thanks,
sugmomo

sugmomo wrote:Thanks Guru


Thanks goyalsau. Can you please elloborate how you arrived to the following step

1 - 1/2 + 1/2 - 3/4 + 3/4 - 5/8 + 5/8 - 11/ 16 .......... Last two terms will be 341 / 512 - 681 / 1024


Thanks,
sugmomo

In such series , when there is some pattern ( like in this one we have the power of 2 in the base with increasing order ) like in this one.


1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/ 64 ........................... - 1/1024

2^1 , 2 ^ 2 , 2 ^ 3 ............ 2 ^ 10


try to form a combination in which you left with two terms. ( i am sorry Buddy Even i don't have clear understanding of this Sometimes i able to do it, Sometimes i d'ont, Like in this one i able to get the first two three terms of the series but then just by looking at the first two three terms you can formulate the last term but i was not able to do it, That's why i have to write all the terms. )

IF Guru can give us some technique then it will be helpful for both of us.