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For every integer k from 1 to 10, inclusive, the kth term of

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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 23 Dec 2012, 23:39
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fozzzy wrote:
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...


I find the GP formula quite handy.

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Re: Numbers , Squences , Indices [#permalink] New post 22 Apr 2013, 23:47
First term in the series is 1/2, second -1/4 third 1/8 fourth is -1/16. and likewise till 10th term.

so sum of the series is = 1/2-1/4+1/8-1/16....

so it is essentially the sum of differences between two adjcent terms((1/2-1/4)+(1/8-1/16) till 10th term.

so it will be 1/4 + 1/16 + etc.. we can stop calculating after this.
1/4 = .25 and 1/16 = .0625 and all the subsequent terms will be much less than .0625.

so the sum will be .25+.0625+etc.. = .3abcd

hence its between 1/4 and 1/2

it took me around a minute to solve it. its a sub 1.5 minute question i believe.
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Re: Numbers , Squences , Indices [#permalink] New post 23 Apr 2013, 01:01
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kabilank87 wrote:
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}})

Multiply on both sides by 2:

2S = 1+(\frac{1}{2^{2}}+..\frac{1}{2^{8}})-(\frac{1}{2}+\frac{1}{2^{3}}+..\frac{1}{2^{9}})

Add both the equations:

3S = 1-\frac{1}{2^{10}}

S = \frac{1}{3}-\frac{1}{3*2^{10}}

D.

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Last edited by mau5 on 24 Apr 2013, 11:54, edited 2 times in total.
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Re: Numbers , Squences , Indices [#permalink] New post 23 Apr 2013, 02:57
Zarrolou wrote:
kabilank87 wrote:
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.

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Re: Numbers , Squences , Indices [#permalink] New post 23 Apr 2013, 03:11
A quick way to do this :

the sum of the sequence for k from 1 to 10 is : Sum = 1/2 - 1/(2^2) + 1/(2^3) ....

Notice that ... represents a very small numbers that even substracted or added to the first three terms (1/2 , -1/4 and 1/8), we can neglect it .

Hence, since 1/2-1/4+1/8 is between 1/4 and 1/2 , the sum will be between 1/4 and 1/2 as well.

Answer : D

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Re: Numbers , Squences , Indices [#permalink] New post 23 Apr 2013, 03:13
kabilank87 wrote:

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.


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.

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Re: Numbers , Squences , Indices [#permalink] New post 24 Apr 2013, 11:21
Zarrolou wrote:
kabilank87 wrote:
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 .

Thanks in advance :)

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Re: Numbers , Squences , Indices [#permalink] New post 24 Apr 2013, 11:47
Zarrolou wrote:
TheNona wrote:

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


Perfect! Thanks a lot :)

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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 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.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 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.

Sum=(1/2) + (-1/4) +(1/8) +(-1/16)+.....
=>1/4 +1/16 => 0.25+0.0625 => .3125

SO answer is between 0.25 and 0.5 . option D

Option A,B & E can be rejected just by looking terms as they are too big or small enough.

Proper calculation approach:
even though u calculate each term
Sum=1/4+1/16+1/64+1/256+1/1024 =>0.25+0.0625+0.015625+0.00390625 +0.000976 =0.3330
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Exponents / powers (2) [#permalink] New post 13 Mar 2014, 07:17
How can I approach this question?

(-1)k+1(½k). T is the sum of the first 10 k, is t
a. > 2
b. between 1 and 2
c. between ½ and 1
d. between ¼ and ½
e. < ¼
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Re: Exponents / powers (2) [#permalink] New post 13 Mar 2014, 07:42
Expert's post
chrish06 wrote:
How can I approach this question?

(-1)k+1(½k). T is the sum of the first 10 k, is t
a. > 2
b. between 1 and 2
c. between ½ and 1
d. between ¼ and ½
e. < ¼


Merging similar topics. please refer to the discussion above.

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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 11 Apr 2014, 23:02
Sequence will be 1/2-1/4+1/8-1/16.......

Sum of first pair= 1/4 = 25%
Sum of 2nd pair=1/16 = Appropriately 6%

So we can see that % is reducing. Therefore next three percentage each of which will not be more than 6%

So, Sum= 25%+6%+ ........ = in between (25 and 50)%
D shows that in percentage.
Re: For every integer k from 1 to 10, inclusive, the kth term of   [#permalink] 11 Apr 2014, 23:02
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