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Founder  Joined: 18 Apr 2015
Posts: 7312
Followers: 122

Kudos [?]: 1427 , given: 6562

Expert's post 00:00

Question Stats: 27% (00:26) correct 72% (00:33) wrong based on 93 sessions

n is an integer.

 Quantity A Quantity B $$(-1)^{2}^{n+1}$$ $$(1)^n$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

_________________
Director Joined: 03 Sep 2017
Posts: 520
Followers: 1

Kudos [?]: 361 , given: 66

Re: n is an integer. [#permalink]
Whatever is n, 2n+1 is always odd. Thus, quantity A is always -1, while quantity B is always 1 because it does not matter if the exponent is even or odd.

How can it be D? What am I losing?
Founder  Joined: 18 Apr 2015
Posts: 7312
Followers: 122

Kudos [?]: 1427 , given: 6562

Re: n is an integer. [#permalink]
Expert's post
Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT $$2n + 1$$ but it is (if you notice very carefully) that 2 (in the first quantity) is $$2^{n+1}$$
_________________ Director Joined: 03 Sep 2017
Posts: 520
Followers: 1

Kudos [?]: 361  , given: 66

Re: n is an integer. [#permalink]
1
KUDOS
Carcass wrote:
Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT $$2n + 1$$ but it is (if you notice very carefully) that 2 (in the first quantity) is $$2^{n+1}$$

Ok, yes the forum wasn't showing it right.

Thanks

I edit my answer then. The exponent in quantity A can be rewritten as $$2^{n+1} = 2n+2$$. Thus, the exponent is always even and -1 to an even exponent becomes 1, so the answer would be C, not D. Sorry for bothering you but may you provide the OE?
Intern Joined: 12 Nov 2017
Posts: 29
Followers: 1

Kudos [?]: 11 , given: 32

Re: n is an integer. [#permalink]
Shouldnt the answer be "B" ?
becuase A is always -1 and B is always 1
Carcass wrote:

n is an integer.

 Quantity A Quantity B $$(-1)^2^^{n+1}$$ $$(1)^n$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Intern Joined: 12 Nov 2017
Posts: 29
Followers: 1

Kudos [?]: 11 , given: 32

Re: n is an integer. [#permalink]
Thank you now I get it
IlCreatore wrote:
Carcass wrote:
Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT $$2n + 1$$ but it is (if you notice very carefully) that 2 (in the first quantity) is $$2^{n+1}$$

Ok, yes the forum wasn't showing it right.

Thanks

I edit my answer then. The exponent in quantity A can be rewritten as $$2^{n+1} = 2n+2$$. Thus, the exponent is always even and -1 to an even exponent becomes 1, so the answer would be C, not D. Sorry for bothering you but may you provide the OE?
Intern Joined: 14 Jun 2018
Posts: 36
Followers: 0

Kudos [?]: 7 , given: 100

Re: n is an integer. [#permalink]
Carcass wrote:

n is an integer.

 Quantity A Quantity B $$(-1)^{2}^{n+1}$$ $$(1)^n$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Carcass, you are still writing it (D)? I agree that it's (C).
Intern Joined: 15 May 2018
Posts: 38
Followers: 0

Kudos [?]: 5 , given: 1

Re: n is an integer. [#permalink]
Carcass wrote:

n is an integer.

 Quantity A Quantity B $$(-1)^{2}^{n+1}$$ $$(1)^n$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given. Re: n is an integer.   [#permalink] 09 Jul 2018, 18:32
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