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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
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Joined: 26 Mar 2017
Posts: 123
Re: If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to  [#permalink]

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New post 22 Jun 2017, 09:03
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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New post 23 Jul 2017, 18:57
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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New post 26 Sep 2018, 10:56
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. \(2^{(2n+1)}\)


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

The answer is D.


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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New post 30 Sep 2018, 08:15
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. *****

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

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

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

= \(2^{-2n} * 2\)

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

The best answer is D.
GMAT Club Bot
If n is a positive integer, then (-2^n)^{-2} + (2^{-n})^2 is equal to &nbs [#permalink] 30 Sep 2018, 08:15

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