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For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(-1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?

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

S = (\frac{1}{2}+\frac{1}{2^{3}}+..\frac{1}{2^{9}}) - (\frac{1}{2^{2}}+..\frac{1}{2^{10}})

For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(-1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?

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

I answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ?

Here we need to find a pattern \frac{1}{2},-\frac{1}{4},\frac{1}{8},... as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms. We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us 1/4, the second pair is 1/8-1/16 positive so we add value to 1/4, so the sum will be greater.(this is true also for the next pairs, so we add to 1/ 4 a positive value for each pair) D

Hope its clear, let me know

Hi Zarro ,

Very clear. But how to approach this kind of problems in GMAT without taking much time. Even though you explanation look very simple and time saving , i am not sure how i respond to the question same way as you explained. Do you have any siggestions how can i approach these problems ?

Thanks in advance.
_________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

Very clear. But how to approach this kind of problems in GMAT without taking much time. Even though you explanation look very simple and time saving , i am not sure how i respond to the question same way as you explained. Do you have any siggestions how can i approach these problems ?

Thanks in advance.

Hi kabilank87,

when we deal with a series (as in this case) the first and most important thing to do is find a pattern.

One you've found that you can continue the series with no limit ( the GMAT will never ask you the exact value of a seires such this one), but the role of patterns is crucial also in question where you're asked to find the N^t^h term of a sequence.
_________________

It is beyond a doubt that all our knowledge that begins with experience.

For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(-1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?

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

I answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ?

Here we need to find a pattern \frac{1}{2},-\frac{1}{4},\frac{1}{8},... as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms. We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us \frac{1}{4}, the second pair is \frac{1}{8}-\frac{1}{16}positive so we add value to \frac{1}{4}, so the sum will be greater.(this is true also for the next pairs, so we add to \frac{1}{4} a positive value for each pair) D

Hope its clear, let me know

Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot .

Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot .

Thanks in advance

The pattern: \frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},...

We have to sum those elements so: \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+... The first term is \frac{1}{2}, to this we subtract 1/4, to the result we add 1/8, and so on As you see the operations involve smaller and smaller term each time. The first thing to notice here is that the sum will be <1/2, we can easily see this: \frac{1}{2}-\frac{1}{4}=\frac{1}{4} and the operations will not produce a result >1/2. Hope it's clear here: the numbers decrease too rapidly to produce a result as big as the first term!

Now we are left with D and E: the only 2 option which result is <1/2. And the question is: will the sum be less than 1/4? We have to find an easy way to see this, consider this fact: \frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},... take the sum of couple of terms: 1st with 2nd, 3rd with 4th, and so on... The result will be positive for each couple, lets take a look:\frac{1}{2}-\frac{1}{4}=\frac{1}{4} for the first one, +\frac{1}{8}-\frac{1}{16}=\frac{1}{16}(>0) and so on.

The thing to take away here is: 1/4+(num>0)+(num>0)+... will NOT be less than 1/4, how could it be if all numbers are positive?

So the sum will be GREATER than 1/4 and LESSER than 1/4.

Hope everything is clear now, I have been as exhaustive as possible, let me know
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot .

Thanks in advance

The pattern: \frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},...

We have to sum those elements so: \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+... The first term is \frac{1}{2}, to this we subtract 1/4, to the result we add 1/8, and so on As you see the operations involve smaller and smaller term each time. The first thing to notice here is that the sum will be <1/2, we can easily see this: \frac{1}{2}-\frac{1}{4}=\frac{1}{4} and the operations will not produce a result >1/2. Hope it's clear here: the numbers decrease too rapidly to produce a result as big as the first term!

Now we are left with D and E: the only 2 option which result is <1/2. And the question is: will the sum be less than 1/4? We have to find an easy way to see this, consider this fact: \frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},... take the sum of couple of terms: 1st with 2nd, 3rd with 4th, and so on... The result will be positive for each couple, lets take a look:\frac{1}{2}-\frac{1}{4}=\frac{1}{4} for the first one, +\frac{1}{8}-\frac{1}{16}=\frac{1}{16}(>0) and so on.

The thing to take away here is: 1/4+(num>0)+(num>0)+... will NOT be less than 1/4, how could it be if all numbers are positive?

So the sum will be GREATER than 1/4 and LESSER than 1/4.

Hope everything is clear now, I have been as exhaustive as possible, let me know

For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(-1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?

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

I answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ?

Here we need to find a pattern \frac{1}{2},-\frac{1}{4},\frac{1}{8},... as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms. We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us \frac{1}{4}, the second pair is \frac{1}{8}-\frac{1}{16}positive so we add value to \frac{1}{4}, so the sum will be greater.(this is true also for the next pairs, so we add to \frac{1}{4} a positive value for each pair) D

Hope its clear, let me know

Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot .

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
11 Oct 2013, 12:22

Since I solved it from a different method than mentioned here, thought it to share.

I guess we all might have deudced there are 10 terms and alternately positive and negative.

I tried with GP sum formula and got lost in calculation.

Since we have alternately + - we can make use of it.

Take 1st term , 2nd term 1/2 and -1/4, add them to get 1/4 Similarly 3rd and 4rth term gives you 1/8 ( 1/8 + -1/16) 1/16 We see a multiplication pattern of 4 here so no need to calculate further.

1/4, 1/16, 1/64, 1/256, 1/1024

Add them to get 256+64+16+4+1= 341/1024

Clealry less than half so lies b/w 1/4 and 1/2

Not a very great method but I guess helps me avoid calculcation mistake if I go for GP sum.

Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
28 Feb 2014, 22:18

Question is written in such a way that it is difficult to comprehend but once it is then it is just a matter of few seconds to crack.Here my answer.

T(K) = (−1)^k+1 * 1/2^k

So, T(1) =1/2 T(2)= -1/4 T(3)= 1/8 T(4)= -1/16 .... and so on No need to calculate higher terms because they doesn't produce any significant increase in sum as answers are widely distributed.