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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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Difficulty:   55% (hard)

Question Stats: 62% (01:44) correct 38% (01:30) wrong based on 576 sessions

HideShow timer Statistics 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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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5
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...

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GRE 1: Q169 V154 Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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Here is an easy way to solve this up
Put N =1
and push the 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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-2^(-2n) + 2^(-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^(1 - 2n)

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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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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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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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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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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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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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Joined: 02 Sep 2009
Posts: 55804
Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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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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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.
wonder what this place will look like without Bunuel
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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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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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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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6

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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GMAT 1: 690 Q48 V35 Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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Vyshak wrote:
-2^(-2n) + 2^(-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^(1 - 2n)

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GRE 1: Q169 V154 Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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FasuSek wrote:
Vyshak wrote:
-2^(-2n) + 2^(-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^(1 - 2n)

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.

Regards
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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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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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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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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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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. $$2^{(2n+1)}$$

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

_________________ Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to   [#permalink] 30 Apr 2017, 04:07

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