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

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04 Mar 2012, 16:18

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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.

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 \(sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3}\).

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

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19 Jun 2012, 05:09

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. _________________

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.

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. _________________

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

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04 May 2013, 06:06

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".

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

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05 May 2013, 09:45

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 \(\frac{-1}{3}\). _________________

If you shut your door to all errors, truth will be shut out.

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

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17 Jul 2014, 10:24

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

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24 Oct 2015, 06:09

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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