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# integral part of x = an integer n such that n <= x <

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Manager
Joined: 25 Jan 2004
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integral part of x = an integer n such that n <= x < [#permalink]  01 Feb 2004, 02:23
[x] = integral part of x = an integer n such that n <= x < (n+1)

Find [1/3] + [2/3] + [(2^2)/3] + [(2^3)/3] + ... + [(2^1000)/3].
GMAT Instructor
Joined: 07 Jul 2003
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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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(2^1001-2)/3 - 500

Can you confirm?
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Joined: 25 Jan 2004
Posts: 92
Location: China
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Kudos [?]: 2 [0], given: 0

I dont have the answer, but the following is what I've got

[(2^n)/3] = (-1/2)+(1/6)*(-1)^n +(1/3)*2^n

Director
Joined: 03 Jul 2003
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AkamaiBrah wrote:
(2^1001-2)/3 - 500

Can you confirm?

I got (2^1001) / 3.

Could you explain why you are subtracting 500.
Manager
Joined: 25 Jan 2004
Posts: 92
Location: China
Followers: 1

Kudos [?]: 2 [0], given: 0

BTW, the answer should be an integer. But 2^1001/3 is not.
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 70 [0], given: 0

Zhung Gazi wrote:
I dont have the answer, but the following is what I've got

[(2^n)/3] = (-1/2)+(1/6)*(-1)^n +(1/3)*2^n

If you look at your answer closely, you will see that it is the same as mine.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 70 [0], given: 0

AkamaiBrah wrote:
(2^1001-2)/3 - 500

Can you confirm?

I got (2^1001) / 3.

Could you explain why you are subtracting 500.

Simplest way to solve this is to assume geometry series of 1001 terms.

If you examine terms, you will notice that rounding every odd and even term subtracts 1/3 and 2/3 respectively. Hence, the answer is the sum of the series or 1/3*(2^1001-1) less the sum of the rounding 500(1/3+2/3) + 1/3 which simplifies to the answer I gave.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Director
Joined: 03 Jul 2003
Posts: 654
Followers: 2

Kudos [?]: 32 [0], given: 0

AkamaiBrah wrote:
Can you confirm?
I got (2^1001) / 3.

Could you explain why you are subtracting 500.

Simplest way to solve this is to assume geometry series of 1001 terms.

I've interpreted [X] as just ().

[x] = integral part of x = an integer n such that n <= x < (n+1)
is the key !

Thanks Akamai
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