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

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink]
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 \(sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3}\).

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

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink]
19 Jun 2012, 05:00

1

This post received KUDOS

Expert's post

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.

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink]
30 Oct 2012, 22:43

Expert's post

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. _________________

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

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink]
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 \(\frac{-1}{3}\). _________________

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]
17 Jul 2014, 09:24

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