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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heshamelaziry
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Brent@GMATPrepNow
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(3 + 3 + 3) + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =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]
(3^2 + 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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heshamelaziry
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Brent Hanneson wrote:(3 + 3 + 3) + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =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]
(3^2 + 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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truplayer256
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ace_gre
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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.
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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heshamelaziry
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