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by pemdas » Sun Feb 26, 2012 5:56 pm
three 3s makes 9 or 3^2
factor out 2*3^2, to get 2*3^2(1+3+3^2+3^3+3^4+3^5)=2*3^2(4+9+27+81+243)
3^2 + 2*3^2(364)= 3^2(1+728)=3^2*(243*3) or 3^2*(3^6)=3^8
dlencz wrote:3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =


(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30
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by Anurag@Gurome » Sun Feb 26, 2012 7:17 pm
dlencz wrote:3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =


(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30
We can write 3 + 3 + 3 = 3²
= 3² + 2 * 3² + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
= 3² + 2 * 3²(1 + 3 + 3² + 3^3 + 3^4 + 3^5)
1 + 3 + 3² + 3^3 + 3^4 + 3^5 is a geometric series and sum to the nth term = a[r� - 1]/(r - 1) = 1(3^6 - 1)/(3 - 1) = 728/2 = 364
So, 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
= 3² + 2 * 3² * 364
= 3² + 3² * 728 = 3²(1 + 728) = 3² * 729 = 3² * 3^6 = [spoiler]3^8[/spoiler]

The correct answer is B.
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by LalaB » Mon Feb 27, 2012 2:41 am
3+ 3 + 3 + 2 X 3^2 + 2 X 3^3 + 2 X 3^4 + 2 X 3^5 + 2 X 3^6 + 2 X 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*3^3+2*3^4 + 2*3^5 + 2*3^6 + 2*3^7 =

here stop! no need to do anything.

from here u can see that previous numbers are the sum of the next number. so at the end, u will have 3^8 (since ur last number is 2*3^7, and the sum of previous numbers will be 3^7. if u dont believe me, write them out)
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