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

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Manager
Joined: 26 Mar 2017
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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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22 Jun 2017, 10:03
Another approach:

(-2^n)^-2 = (2^-n)^2

(-1)^-2 * 2^-2n + 2^-2n

= 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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23 Jul 2017, 19:57
Hello,

Could anybody help me? Why is this wrong:

$$(-2^{n})^{-2}$$ = $$((-2)^n)^{-2}$$ = $$-2^{-2n}$$
$$(2^{-n})^{2}$$ = $$((2)^{-n})^{2}$$ = $$2^{-2n}$$

$$-2^{-2n}$$ + $$2^{-2n}$$ = 0

The answer I got was 0. I can follow the steps to get to D, but I'm not seeing my mistake, which originally led me to A.

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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26 Sep 2018, 11:56
broall 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. $$2^{(2n+1)}$$

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

Can someone explain how 2/2^2n become 1/2^(2n-1) please?

=\frac{2}{2^{2n}}\\
=\frac{1}{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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30 Sep 2018, 09:15
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. $$2^{(2n+1)}$$

*** negative base having positive even exponent is equal to positive base having even exponent. No difference. *****

$$(-2^{n})^{-2}$$ +$$(2^{-n})^{2}$$

= $$2^{-2n} + 2^{-2n}$$

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

= $$2^{-2n} * 2$$

= $$2^{-2n +1}$$

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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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05 Jun 2019, 11:39
You can't factor out a common number if it's not the same base.

$$(-2^{n})^{-2} + (2^{-n})^{2}$$

$$(-1^{n}*2^{n})^{-2} + 2^{(-n*2)}$$

$$(-1^{-2n}*2^{-2n}) + 2^{-2n}$$

$$(\frac{1}{-1^{2n}}*2^{-2n}) + 2^{-2n}$$
Base 1 to the power of anything is 1 and even exponents remove negative...

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

$$2^{-2n} + 2^{-2n}$$

$$2^{-2n}(1+1)$$

$$2^{-2n}(2)$$

$$2^{-2n+1}$$
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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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10 Jul 2019, 16:51
Got this wrong on EP2, GMATPrep 6. It was question 16.

STUPID mistakes were made as I took the algebraic approach and incorrectly didn't apply the even power to the base properly.

The plug-in method is so much more effective on this.
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Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to   [#permalink] 10 Jul 2019, 16:51

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