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Exponents

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by heshamelaziry » Sat Nov 14, 2009 10:29 pm
3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =

[spoiler]OA 3^8. I understand the first 3 terms will be 3^2. but can't go afterward ?[/spoiler]
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Re: Exponents

by Brent@GMATPrepNow » Sun Nov 15, 2009 8:11 am
heshamelaziry wrote:3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =

[spoiler]OA 3^8. I understand the first 3 terms will be 3^2. but can't go afterward ?[/spoiler]
(3 + 3 + 3) + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^2 + 2 * 3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^3) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3*3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^4 + 2 * 3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^5) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3^6 + 2 * 3^6 + 2 * 3^7 =
etc
3^8
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Re: Exponents

by heshamelaziry » Sun Nov 15, 2009 12:19 pm
Brent Hanneson wrote:
heshamelaziry wrote:3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =

[spoiler]OA 3^8. I understand the first 3 terms will be 3^2. but can't go afterward ?[/spoiler]
(3 + 3 + 3) + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^2 + 2 * 3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^3) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3*3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^4 + 2 * 3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^5) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3^6 + 2 * 3^6 + 2 * 3^7 =
etc
3^8

How did (3^2 + 2 * 3^2) become
(3*3^2) ?
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by truplayer256 » Sun Nov 15, 2009 1:39 pm
How did (3^2 + 2 * 3^2) become
(3*3^2) ?
Factor out a 3^2 to get:

3^2(1+2)=3(3^2) or 3^3.
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by ace_gre » Thu Dec 03, 2009 12:49 pm
Another approach to this problem. Treat the text in blue as GP..

3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7
Simplify the blue portion,
2*3^1 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7.

Term t1=2*3^1=2*3
Common ratio(r)= t2/t1=2 * 3^2/2*3^1=3
Sum S= t1((r^n) - 1)/(r-1)
S=(2*3) * ((3^7)-1)/(3-1)
S=3*((3^7)-1)


Total series sum = 3+S=3+3*((3^7)-1) = 3(1+(3^7)-1)=3*3^7=3^8.
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by heshamelaziry » Thu Dec 03, 2009 12:56 pm
exactly what i tried to do, but failed to figure out.
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