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2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = ? I realise there is a

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Intern
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Joined: 31 May 2006
Posts: 44

Kudos [?]: 11 [0], given: 0

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = ? I realise there is a [#permalink]

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New post 15 Jul 2006, 16:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = ?

I realise there is a trick involved, but I can't seem to figure this one out...answer is 2^9.




SQ

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CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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New post 15 Jul 2006, 16:18
Yes there is a trick that will make the calculation really easy and you can solve in just 20-30 seconds.

Take first two terms at a time
(2+2)+2^2+2^3+2^4+2^5+2^6+2^7+2^8
= (2^2+2^2)+2^3+2^4+2^5+2^6+2^7+2^8
= (2^3+2^3)+2^4+2^5+2^6+2^7+2^8
= (2^4+2^4)+2^5+2^6+2^7+2^8
= (2^5+2^5)+2^6+2^7+2^8
= (2^6+2^6)+2^7+2^8
= (2^7+2^7)+2^8
= 2^8 + 2^8
= 2^9 ANSWER
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Senior Manager
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Joined: 07 Jul 2005
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Location: Sunnyvale, CA
Re: GMAT prep [#permalink]

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New post 15 Jul 2006, 16:31
shehreenquayyum wrote:
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = ?

I realise there is a trick involved, but I can't seem to figure this one out...answer is 2^9.




SQ


Use the formula for G.P.
Sn = a * (r^n - 1) / (r -1)

1 + (2^0 +2^1 ... + 2^8)
= 1 + (1.(2^9 - 1)/1
= 1 + 2^9 - 1
= 2^9

Kudos [?]: 14 [0], given: 0

Senior Manager
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Joined: 12 Mar 2006
Posts: 363

Kudos [?]: 77 [0], given: 3

Schools: Kellogg School of Management
 [#permalink]

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New post 16 Jul 2006, 06:52
sum of n terms of a gp is a(1 - r^n+1)/(1-r)

a - first term
r - rate of progression

here 2(1-2^9)/-1 +2 = sum = 2^9

Kudos [?]: 77 [0], given: 3

  [#permalink] 16 Jul 2006, 06:52
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