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# Powers

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Director
Joined: 10 Feb 2006
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17 Sep 2006, 17:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Can you guys solve this.
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GMAT the final frontie!!!.

SVP
Joined: 05 Jul 2006
Posts: 1747
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17 Sep 2006, 19:15
The best one who approached this problem is dahiya a 700 scorer

and this is the way he did it

2+2 = 2^2

2^2 + 2^2 = 2^3

2^3+2^3 = 2^4

2^4+2^4 = 2^5

....

.....
Senior Manager
Joined: 28 Aug 2006
Posts: 304
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17 Sep 2006, 21:00
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.

Hey this is simple if u know the formaula to calculate the sum of n terms of a Geometric Progression.
Excluding the first 2 in the series the remaining series is a GP with 8 terms and common ratio of 2.

So the formula of a GP whose first term is a with common ratio r and number of terms n is

a(r^n-1)/r-1 (if r>1)

a(1-r^n)/1-r (if r<1).

Here excluding the first 2 the remaining series is
2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8.
here a=2 r=2 and n=8
So sum is 2(2^8-1)/2-1 = 2 (256-1)= 510.

Adding the first two the sum has to be 510+2 = 512 = 2^9
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Last edited by cicerone on 25 Sep 2008, 01:06, edited 1 time in total.
Director
Joined: 06 May 2006
Posts: 791
Followers: 3

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17 Sep 2006, 23:02
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.

Sum of GP where a = 2, r = 2, n = 8

a*(r^n - 1)/(r - 1) = 2*(2^8 - 1)/(1) = 255*2 = 510

Then we have 2 + Sum = 2 + 510 = 512 = 2^9.
_________________

Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

Director
Joined: 28 Dec 2005
Posts: 752
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17 Sep 2006, 23:08
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.

This is straight binary math....

2+4+8+16+32+64+128+256 = 510

Re: Powers   [#permalink] 17 Sep 2006, 23:08
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