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Director
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Powers [#permalink] New post 17 Sep 2006, 16:14
= 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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 [#permalink] New post 17 Sep 2006, 18:15
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

....

.....
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Re: Powers [#permalink] New post 17 Sep 2006, 20:00
alimad wrote:
= 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, 00:06, edited 1 time in total.
Director
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Re: Powers [#permalink] New post 17 Sep 2006, 22:02
alimad wrote:
= 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.
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Re: Powers [#permalink] New post 17 Sep 2006, 22:08
alimad wrote:
= 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

add 2, 512..
Re: Powers   [#permalink] 17 Sep 2006, 22:08
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