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\(2+2=2*2=2^2\) (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.

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