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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? [#permalink] New post 13 Aug 2010, 09:15
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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

A. 2^9
B. 2^10
C. 2^16
D. 2^35
E. 2^37
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Apr 2012, 00:01, edited 4 times in total.
Edited the question and added the OA
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Re: GMAT prep Q [#permalink] New post 14 Aug 2010, 09:18
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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

A. 2^9
B. 2^10
C. 2^16
D. 2^35
E. 2^37

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)=2+\frac{2*(2^{8}-1)}{2-1}=2^9.

Answer: A.

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new [#permalink] New post 01 Apr 2012, 13:24
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there are nine numbers, so group them up in 3.

[2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]

[2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7

2^3*64=2^3*2^6=2^9

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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? * 2^9 * [#permalink] New post 01 Apr 2012, 18:17
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amitdgr wrote:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

* 2^9
* 2^10
* 2^16
* 2^35
* 2^37


See for pattern:
2 + 2 =2^2
2^2+2^2=2^3
2^3 + 2^3=2^4
........
2^8 + 2^8=2^9

wouldnot take more than 30 seconds.

hope this helps..!!

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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? * 2^9 * [#permalink] New post 01 Apr 2012, 23:46
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cangetgmat wrote:
I was stuck performing addition of GP.
+1 to you


You can do it using sum of GP as well.

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.

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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? [#permalink] New post 04 Apr 2012, 00:03
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? [#permalink] New post 07 Sep 2013, 09:26
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?   [#permalink] 07 Sep 2013, 09:26
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