Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

04 Mar 2012, 15:18

2

This post received KUDOS

14

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

59% (01:57) correct 41% (01:42) wrong based on 234 sessions

HideShow timer Statistics

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]

Show Tags

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.
_________________

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]

Show Tags

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]

Show Tags

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.