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2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

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by killer1387 » Tue Mar 27, 2012 4:07 pm
NYC493 wrote:2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

Fastest way to simplify please...
2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)
=2+2(1+2+2^2+.......+2^7)
=2+2(2^8-1)=2+2^9-2
=2^9=512
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by Brent@GMATPrepNow » Tue Mar 27, 2012 4:15 pm
NYC493 wrote:2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

Fastest way to simplify please...
A nice pattern develop here.

Add the first 2 terms (2 and 2) to get 2^2
Now add 2^2 to the next term (2^2) to get 2^3
Now add 2^3 to the next term (2^3) to get 2^4
Now add 2^4 to the next term (2^4) to get 2^5
.
.
.
Now add 2^8 to the next term (2^8) to get 2^9

Cheers,
Brent

PS: In general, 2^x + 2^x = 2^(x+1)
Why?
Well, 2^x + 2^x = 2(2^x) [just like k + k = 2k, and m^5 + m^5 = 2m^5]
= (2^1)(2^x)
= 2^(x+1)

Cheers,
Brent
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by pemdas » Tue Mar 27, 2012 5:17 pm
NYC493 wrote:2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

Fastest way to simplify please...
this is simple series problem, a*r^0+a*r^1+...a*r^n, where a=2 and r=1->7. To hide the series sequence additional number '2' was used, e.g. 2+2*2^0+(2*2^1)+(2*2^2)+(2*2^3)+(2*2^4)+(2*2^5)+(2*2^6)+(2*2^7)

The sum of series will be 2*[2^(7+1) -1]/(2-1)=2*(2^8 -1) or 2^9 - 2
Since we have additional 2, rewritten formula will be 2^9 - 2 + 2 = 2^9
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by ronnie1985 » Wed Mar 28, 2012 9:40 am
This is a geometric series.
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by amit28it » Sat Mar 31, 2012 12:43 am
NYC493 wrote:2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

Fastest way to simplify please...
The most common and simple method for solving this is
2+2+4+8+16+32+64+128+256= 512
Just add the same number to itself after the first 2.
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