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exponents

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Source: — Problem Solving |

by dmateer25 » Wed Nov 26, 2008 11:10 am
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 =

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

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

2^3(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5)

2^4(2 + 2 + 2^2 + 2^3 + 2^4)

2^5(2 + 2 + 2^2 + 2^3)

2^6(2 + 2 + 2^2 )

2^7(2^2)

2^9
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by resepulv » Wed Nov 26, 2008 12:38 pm
Is equal to: 1+SUM(2^i), i=0,...,8
the formula to solve this is: SUM(a^i), i=0,...,n is equal to: (1-a^(n+1))/(1-a).

So the solution is: 1+(1-2^9)/(1-2)=1+2^9-1=2^9
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by vishubn » Wed Nov 26, 2008 11:35 pm
there is a solution posted by cramya ! in a much comprehensive way ... please search the post :)

Vishu
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!
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