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:

\(2+2=2*2=2^2\) (the sum of first and second terms); \(2^2+2^2=2^3\) (the sum of previous 2 terms and the third term); \(2^3+2^3=2^4\) (the sum of previous 3 terms and the fourth term); ... The same will continue and finally we'll get \(2^8+2^8=2*2^8=2^9\), the sum of previous 8 terms and the 9th term.

Answer: A.

OR: we can identify that the terms after the first one represent geometric progression.

Sum of the terms of geometric progression is given by: \(Sum=\frac{a*(r^{n}-1)}{r-1}\), where \(a\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(>1\).

In our original question we have 2 plus G.P. with 8 terms, so:

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 1 + (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8)\) The terms in brackets make a GP with first term = 1 and common ratio = 2

GP = \(1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\) Sum of 9 terms of GP \(= 1(1 - 2^9)/(1-2) = 2^9 - 1\)

Sum of the required series = 1 + sum of GP = \(1 + 2^9 - 1 = 2^9\)

To check out a discussion on both the methods, check out my blog http://www.veritasprep.com/blog/categor ... er-wisdom/ today evening. My this week's post discusses a question very similar to this one. _________________

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

\(2+2=2*2=2^2\) (the sum of first and second terms); \(2^2+2^2=2^3\) (the sum of previous 2 terms and the third term); \(2^3+2^3=2^4\) (the sum of previous 3 terms and the fourth term); ... The same will continue and finally we'll get \(2^8+2^8=2*2^8=2^9\), the sum of previous 8 terms and the 9th term.

Answer: A.

OR: we can identify that the terms after the first one represent geometric progression.

Sum of the terms of geometric progression is given by: \(Sum=\frac{a*(r^{n}-1)}{r-1}\), where \(a\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(>1\).

In our original question we have 2 plus G.P. with 8 terms, so:

I didn't think of it as a progression, but I started by taking out 2 as a common term. Taking out the first common term of 2 you end up with 2(1+1+2+2^2+...+2^8).

In the end you end up with 2*2*2*2*2*2*2*2*2 = 2^9.

You can either finish the whole process, taking out 2 as a common term in each step, or you will end up noticing that in every step you always end up with 9 numbers (terms of 2), inside or outside the parentheses (1+1 = 2, so it counts as one number). After taking out another term of two, one more number is added in those outside the parenthesis and one number is removed from the ones inside the parentheses. You have 9 terms, so in the end there will be 9 two's outside the parentheses: 2^9.

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

London is the best kept secret of the corporate world. It is English speaking and time delayed by only 5 hours. That means when London goes home at 5...