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If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to

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If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post Updated on: 19 Jul 2016, 21:57
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If n is a positive integer, then \((-2^{n})^{-2}\) +\((2^{-n})^{2}\) =

A. 0

B. \(2^{(-2n)}\)

C. \(2^{(2n)}\)

D. \(2^{(-2n+1)}\)

E. \(2^{(2n+1)}\)

Originally posted by lpetroski on 28 Mar 2016, 10:27.
Last edited by Bunuel on 19 Jul 2016, 21:57, edited 1 time in total.
Edited the question.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 19 Jul 2016, 15:54
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HI devbond,

You really should post this in the PS Forum here:

gmat-problem-solving-ps-140/

That having been said, anyone who is reading this should work off of what is printed in the attachment (as your notation is incorrect).

This question can be solved rather easily by TESTing VALUES.

IF... X = 1, then the value of the calculation becomes...

(-2^1)^(-2) + (2^-1)^2 =
(-2)^(-2) + (1/2)^2 =
1/4 + 1/4 =
1/2

So we're looking for an answer that equals 1/2 when X = 1. There's only one answer that matches...

Final Answer:

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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 15 Apr 2016, 02:36
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General Discussion
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 28 Mar 2016, 10:57
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-2^(-2n) + 2^(-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^(1 - 2n)

Answer: D
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 12 Apr 2016, 04:58
lpetroski wrote:
If n is a positive integer, then \((-2^n)^-^2\) +\((2^-^n)^2\) =

a. 0
b. 2^(-2n)
c. 2^(2n)
d. 2^(-2n+1)
e. 2^(2n+1)


Can someone edit and properly encode that question please?
I'm sure there are moderators in quant forum...
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 12 Apr 2016, 05:10
Nez wrote:
lpetroski wrote:
If n is a positive integer, then \((-2^n)^-^2\) +\((2^-^n)^2\) =

a. 0
b. 2^(-2n)
c. 2^(2n)
d. 2^(-2n+1)
e. 2^(2n+1)


Can someone edit and properly encode that question please?
I'm sure there are moderators in quant forum...


Thanks Nez. I have corrected the question.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 12 Apr 2016, 23:29
Vyshak wrote:
Nez wrote:
lpetroski wrote:
If n is a positive integer, then \((-2^n)^-^2\) +\((2^-^n)^2\) =

a. 0
b. 2^(-2n)
c. 2^(2n)
d. 2^(-2n+1)
e. 2^(2n+1)


Can someone edit and properly encode that question please?
I'm sure there are moderators in quant forum...


Thanks Nez. I have corrected the question.

well done Vyshak, but it's half done.
The options are not properly encoded.

Posted from my mobile device
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 12 Apr 2016, 23:41
Nez wrote:
Vyshak wrote:
Nez wrote:
[quote="lpetroski"]If n is a positive integer, then \((-2^n)^-^2\) +\((2^-^n)^2\) =

a. 0
b. 2^(-2n)
c. 2^(2n)
d. 2^(-2n+1)
e. 2^(2n+1)


Can someone edit and properly encode that question please?
I'm sure there are moderators in quant forum...


Thanks Nez. I have corrected the question.

well done Vyshak, but it's half done.
The options are not properly encoded.

Posted from my mobile device [/quote]
Hi Nez,

I understand your concerns that the options are not encoded using mathematical conventions but they are legible and I feel the options, unlike the question for which editing was required, can be understood by everyone.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 15 Apr 2016, 07:47
Hi Vyshak,

Would you mind giving me administrator status briefly; u tell me what those choices are, then I could properly encode them myself.
I don't want to border you. Just let me do it.
It will be a pleasure.
God bless you as I wait for your acceptance.

Posted from my mobile device
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 15 Apr 2016, 08:00
1
Nez wrote:
Hi Vyshak,

Would you mind giving me administrator status briefly; u tell me what those choices are, then I could properly encode them myself.
I don't want to border you. Just let me do it.
It will be a pleasure.
God bless you as I wait for your acceptance.

Posted from my mobile device


Dear Nez, mathematically everything was correct there. Still edited.

P.S. There is NO such thing as briefly giving administrator status.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 15 Apr 2016, 10:21
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lpetroski wrote:
If n is a positive integer, then \((-2^{n})^{-2}\) +\((2^{-n})^{2}\) =

A. 0

B. \(2^{(-2n)}\)

C. \(2^{(2n)}\)

D. \(2^{(-2n+1)}\)

E. \(2^{(2n+1)}\)


Now that looks good.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 27 May 2016, 08:51
Did a bunch of awful math and got this one wrong on Exam Pack #2, Test #6

On review, I plugged in N=1 and got answer choice D in about 45 seconds.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 19 Jul 2016, 13:15
devbond wrote:
If n is a positive integer, then (-2n)^-2 + (2^-n) is equal to

a)0
b)2^-2n
c)2^n
d)2^-2n+1
e)2^2n+1


Please post with options and OA included.

We know that a^-m = 1/a^m.

(-2^n)-2 ) can be written as 1/(-2^n)^2 => 1/2^2n ( Here a^m * a^n = a^m+n ) ( -2^n * -2^n = 2^2n )

(2^-n)^2 can be written as 1/2^2n.

1/2^2n + 1/2^2n = 2^-2n+1.
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If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 23 Jul 2016, 08:55
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Here's the algebraic solution:



\(\left( -2^{n} \right)^{-2}+\left( 2^{-n} \right)^{2}\; =\; ?\;\)

\(\; as\; a^{-b}=\frac{1}{a^{b}}\; \;\)

\(\left( -2n \right)^{-2}\; +\; ...\; =\; \frac{1}{\left( -2^{n} \right)^{2}}\; +\; ...\;\)

\(as\; every\; positive\; number\; squa\mbox{re}d\; is\; positive\; \left( -2^{n} \right)^{2}\; becomes\; \left( 2^{2n} \right)\)

\(=\; \frac{1}{\left( 2^{2n} \right)}\; +\; \left( 2^{-n} \right)^{2}=\; ...\; +\; \left( 2^{-2n} \right)\; = ...\; +\; \frac{1}{\left( 2^{2n} \right)}=\;\)


\(\frac{1}{2^{2n}}+\frac{1}{2^{2n}}\; =\; \frac{1}{4^{n}}+\frac{1}{4^{n}}=\; 2\cdot 2^{\left( -2n \right)}\; =\; 2^{\left( -2n+1 \right)}\)


\(Answer\; D\)
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 08 Aug 2016, 07:34
Vyshak wrote:
-2^(-2n) + 2^(-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^(1 - 2n)

Answer: D


Why is -2^(-2n) = 1/4^n, could you please expand your answer?
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 08 Aug 2016, 07:51
FasuSek wrote:
Vyshak wrote:
-2^(-2n) + 2^(-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^(1 - 2n)

Answer: D


Why is -2^(-2n) = 1/4^n, could you please expand your answer?


Hey
-2^(-2n) = [1/2]^2n =[1/4]^n
The property used here is => X raised to the power say negative p is one over X raised to the power positive p

For instance

"x^–2" just means "1/x^2


P.S => To make life easier put N=1 in the original question and you would arrive at D option.


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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 15 Aug 2016, 09:34
Hi all,

Isn't (−2^n)^−2 = 2n^2? Since e.g. (-6)^-3 = (6)^3. What am I missing?

Thanks!
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 15 Aug 2016, 09:43
HarveyKlaus wrote:
Hi all,

Isn't (−2^n)^−2 = 2n^2? Since e.g. (-6)^-3 = (6)^3. What am I missing?

Thanks!


\((-6)^{-3}=\frac{1}{(-6)^3}\)

Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 18 Apr 2017, 07:57
Option D

\((-2^n)^{-2} + (2^{-n})^2 = (1/4)^{n} + (1/4)^{n} = (1/2)^{2n} + (1/2)^{2n} = (1/2^{2n - 1})*(1/2 + 1/2) = 1/2^{2n - 1} = 2^{-(2n - 1)}= 2^{-2n + 1}\)
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 30 Apr 2017, 04:07
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lpetroski wrote:
If n is a positive integer, then \((-2^{n})^{-2}\) +\((2^{-n})^{2}\) =

A. 0

B. \(2^{(-2n)}\)

C. \(2^{(2n)}\)

D. \(2^{(-2n+1)}\)

E. \(2^{(2n+1)}\)


\((-2^n)^{-2}+(2^{-n})^2 \\
=\frac{1}{(-2^n)^2}+2^{-2n} \\
=\frac{1}{(-2)^{2n}}+\frac{1}{2^{2n}}\\
=\frac{1}{2^{2n}}+\frac{1}{2^{2n}}\\
=\frac{2}{2^{2n}}\\
=\frac{1}{2^{2n-1}} \\
=2^{-(2n-1)}\\
=2^{-2n+1}
\)

The answer is D.
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