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

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For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 03 Jun 2007, 02:27
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60% (02:37) correct 40% (01:39) wrong based on 95 sessions
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 0.5 and 1

D. between 0.25 and 0.5

E. less than 0.25


[Reveal] Spoiler:
Shall I use the formula for Geometric Progression;

Sn = a*[ (1-r^n)/(1-r) ]

and calculate finally up to Sn = ( 2^10 - 1 )/ (3* (2^10)) = 1023/3*1024

This seems to be such a long and tedious calculation!! and I don't think we need to know formulae for Geometric Progressions etc.

What should be fast approach to solve this problem? Please explain.


OPEN DISCUSSION OF THIS QUESTION IS HERE: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Aug 2013, 05:24, edited 2 times in total.
Added OA.
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 [#permalink] New post 03 Jun 2007, 07:40
LM,
i am afraid but this is what u are asked to do. right method

But u don't need to calculate A= 1023/3*1024. Just A is almost same with
1024/3*1024 = 1/3 (ROUND UP unit digit)

Vote for D
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ANS [#permalink] New post 03 Jun 2007, 11:20
suppose
S=1+1/2+1/4+1/8............................ this is an infinite GP and here
s comes to be S=2(which is max).
Now we are given
S=1/2-1/4+1/8-1/16+...........
here solving for first two terms we get 1/4 i.e .25 means our answer should be greater than .25 as all other terms gives positive in pair of two.
Now the max. value of this expansion will be 1 when all the terms are positive and upto infinite but in this every alternate term is negative thus reducing the sum of expansion to half of 1 that is 1/2(.5) which can be the max. value.
Thus value lies b/w .25 and .50.

D should be the answer.
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Re: ANS [#permalink] New post 03 Jun 2007, 15:29
deepakguptaeng@gmail.com wrote:
suppose
S=1+1/2+1/4+1/8............................ this is an infinite GP and here
s comes to be S=2(which is max).
Now we are given
S=1/2-1/4+1/8-1/16+...........
here solving for first two terms we get 1/4 i.e .25 means our answer should be greater than .25 as all other terms gives positive in pair of two.
Now the max. value of this expansion will be 1 when all the terms are positive and upto infinite but in this every alternate term is negative thus reducing the sum of expansion to half of 1 that is 1/2(.5) which can be the max. value.
Thus value lies b/w .25 and .50.

D should be the answer.



Good explanation. Answer is correct. But please remember it is not "infnite G.P." Up to 10 terms only.
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For every integer k from 1 to 10, inclusive, the kth term... [#permalink] New post 04 Aug 2013, 11:05
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

Last edited by Zarrolou on 04 Aug 2013, 11:07, edited 1 time in total.
Merging similar topics.
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Re: For every integer k from 1 to 10, inclusive, the kth term... [#permalink] New post 04 Aug 2013, 11:24
Expert's post
njkhokh wrote:
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



From the given sum, T \(= \frac{1}{2}-\frac{1}{2^2}.......-\frac{1}{2^{10}}\)

\(=[ \frac{1}{2}+\frac{1}{2^3}]+... -[\frac{1}{2^2}+\frac{1}{2^4}...]\)

\(=[ \frac{1}{2}+\frac{1}{2^3}]+... -\frac{1}{2}[\frac{1}{2}+\frac{1}{2^3}...]\)

\(=\frac{1}{2}[ \frac{1}{2}+\frac{1}{2^3}+. . . . . . +\frac{1}{2^9}...]\) =\(\frac{1}{2}[0.5+0.125+0.0xxx+..] \approx =0.3125.\)

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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 11 Aug 2013, 05:24
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Re: For every integer k from 1 to 10, inclusive, the kth term of   [#permalink] 11 Aug 2013, 05:24
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