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For every integer m from 1 to 100, inclusive, the mth term

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For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 04 Mar 2012, 15:18
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For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the fi…rst 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.

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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 04 Mar 2012, 16:24
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enigma123 wrote:
For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the fi…rst 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.


1st term \((-1)^1*2^{-1}=-\frac{1}{2}\);
2nd term \((-1)^2*2^{-2}=\frac{1}{4}\);
3rd term \((-1)^3*2^{-3}=-\frac{1}{8}\);
4th term \((-1)^4*2^{-4}=\frac{1}{16}\);
5th term \((-1)^5*2^{-5}=-\frac{1}{32}\);
6th term \((-1)^6*2^{-6}=\frac{1}{64}\);
...

The sum of 1st and 2nd = \(-\frac{1}{4}\);
The sum of 3rd and 4th = \(-\frac{1}{16}\);
The sum of 5th and 6th = \(-\frac{1}{64}\);
...

So, we should basically sum 100/2=50 terms which form a geometric progression with the first term of \(-\frac{1}{4}\) and common ratio of \(\frac{1}{4}\): \(-\frac{1}{4}\), \(-\frac{1}{16}\), \(-\frac{1}{64}\), ...

The sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.

So, in our case, since the # of terms is large enough, the sum will be very close to \(sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3}\).

Answer: C.
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 05 May 2012, 04:40
I don't get it. Isn't the first term -1/2 and the common ratio 1/2?
And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.?
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 19 Jun 2012, 04:09
Stiv wrote:
I don't get it. Isn't the first term -1/2 and the common ratio 1/2?
And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.?


Well both the solution would fetch you the same result of -1/3 if you consider it as an infinite series..
and that would solve the question.
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 19 Jun 2012, 05:00
1
enigma123 wrote:
For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the fi…rst 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.


I have discussed this question in details using both the methods - GP perspective and non GP perspective.
Check it out here:
http://www.veritasprep.com/blog/2012/04 ... rspective/
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 30 Oct 2012, 22:37
Why should we consider this as an infinite sequence?
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 30 Oct 2012, 22:43
Ousmane wrote:
Why should we consider this as an infinite sequence?


We are not saying that it's an infinite sequence. But if it were, then the sum would be -1/3 and since we have the sum of the large enough number of terms (100), then the actual sum would be very close to -1/3.
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 04 May 2013, 05:06
Ousmane wrote:
Why should we consider this as an infinite sequence?



Lets say we have only the first term m=1 , then the sum of the series (since it is just one term) , would be -1/2
Now , as explained by Bunuel , lets suppose there are infinite terms , in which case the sum is -1/3 . Hence sum of 1 to 100 terms SHOULD lie between these two range "-1/2 and -1/3".

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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 05 May 2013, 08:45
A general term of the expression is:-
\((-1)^m.2^{-m}\)
Hence, the 1st 100 terms can be written as:-
\(t_1=\frac{-1}{2}\)
\(t_2=\frac{1}{4}\)
\(t_3=\frac{-1}{8}\)
.
.
.
.
.
\(t_{100}=\frac{1}{2^{100}}\)
Now, we know that the sum of the first "n" terms of a Geometric progression with a common ratio r<1 is:
\(\frac{a(1-r^n)}{1-r}\)
where a=1st term, r=common ratio and n=no. of terms.
Now for this case the expression would be(after putting all the values):-
\(N=\frac{\frac{-1}{2}[1-(\frac{-1}{2})^{100}]}{1-(\frac{-1}{2})}\)
\(=\frac{\frac{-1}{2}[1-\frac{1}{2^{100}}]}{\frac{3}{2}}\)
\(=\frac{-1}{3}.[1-\frac{1}{2^{100}}]\)
Now, since \(2^{100}\) is a very large number compared to 1,
Hence, we can cay that \(1-(\frac{1}{2^{100}})\) is slightly less than 1
Hence, N is slightly less than \(\frac{-1}{3}\).
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 22 Nov 2017, 06:58
Bunuel Why don't we use the formula a(r^n-1)/r-1?
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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New post 22 Nov 2017, 07:32
radhikakhemka007 wrote:
Bunuel Why don't we use the formula a(r^n-1)/r-1?


We can use this formula but the result will be same i.e. -1/3
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Re: For every integer m from 1 to 100, inclusive, the mth term &nbs [#permalink] 22 Nov 2017, 07:32
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For every integer m from 1 to 100, inclusive, the mth term

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