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

Question

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)}

DensetsuNo · Accepted Answer

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

stonecold · Answer

Here is an easy way to solve this up 
Put N =1 
and push the D option...

EMPOWERgmatRichC · Answer

HI devbond,

You really should post this in the PS Forum here:

https://gmatclub.com/forum/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} + (\frac{1}{2})^2 = $
$\frac{1}{4} + \frac{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:

Show Spoiler

D

GMAT assassins aren't born, they're made,
Rich

broall · Answer

$(-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.

Kurtosis · Answer

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

Answer: D

HarveyKlaus · Answer

Hi all,

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

Thanks!

Bunuel · Answer

(-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 All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39 All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

RR88 · Answer

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}

jabhatta2 · Answer

Hi  Bunuel   zhanbo  -- Wanted to focus on the first expression (marked in yellow)

When I inserted   into the yellow function : I got ( -1/4 )

But that is clearly wrong as I should be getting   if i insert n = 1 into the yellow expression.

Here is how I got ( - 1 / 4)

Could you please let me know, where is the mistake in my step by step ?

Bunuel · Answer

The point is that  (-2^{n})^{-2}  means  (-1*2^{n})^{-2} . So, the minus sign there just means "multiplied by -1" and the exponent of -2 should be applied to -1 too. Thus, if we go the way you are doing we'd get:

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

Now, if you plug 1 for n, you'd' get 1/4.

Here is how I'd solve this question.

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)}

Because -2 is even, then  (-2^{n})^{-2}  is the same as  (2^{n})^{-2} . So, we have:

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

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

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

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

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

Answer: D.

Hope it helps.

arnavghatage · Answer

Bunuel 
The equation becomes -2^-2n + 2^-2n.
Why can't we take -2^-2n common?
-2^-2n(1-1)

So this becomes 0.

Krunaal · Answer

You made an error while simplification of  (-2^n)^{-2}

(-2^n)^{-2} => (-1*2^n)^{-2} => (-1^{-2} * 2^{-2n}) = 2^{-2n}

Now we have  2^{-2n}  +  2^{-2n}  = 2*( 2^{-2n} ) =   2^{1-2n}

Answer D.

egmat · Answer

This problem can feel tricky with all those negative exponents and parentheses flying around, but let me walk you through it step by step. The key is to handle each term carefully and then spot the pattern that makes everything click.

Let's break this down systematically:

Step 1: Simplify the first term  (-2^n)^{-2}

Think about what we have here. We're taking  (-2^n)  and raising it to the power of  -2 .

Here's what you need to see:  (-2^n)  means  -(2^n) , so we can rewrite this as:
  ^{-2}

Now, when you raise a negative number to an even power, the negative disappears. So:
  ^{-2} = (2^n)^{-2} 	imes (-1)^{-2}

Since  (-1)^{-2} = \frac{1}{(-1)^2} = \frac{1}{1} = 1 , we get:
 (2^n)^{-2} = 2^{-2n}

Step 2: Simplify the second term  (2^{-n})^2

This one's more straightforward. We're squaring  2^{-n} .

Using the power of a power rule (multiply the exponents):
 (2^{-n})^2 = 2^{(-n) 	imes 2} = 2^{-2n}

Step 3: Combine the terms

Notice how both terms simplified to exactly the same thing:  2^{-2n}

So we're adding:
 2^{-2n} + 2^{-2n} = 2 	imes 2^{-2n}

Step 4: Express as a single power

Let's rewrite  2  as  2^1 :
 2^1 	imes 2^{-2n}

When multiplying powers with the same base, add the exponents:
 2^{1 + (-2n)} = 2^{1-2n} = 2^{-2n+1}

Answer: D

The complete solution on Neuron shows you the systematic framework for handling all negative exponent problems, plus common trap patterns to watch out for and time-saving recognition techniques. You can check out the  detailed step-by-step solution on Neuron by e-GMAT  to master the underlying patterns systematically. You can also explore other GMAT official questions with comprehensive solutions on Neuron for structured practice  here .

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

Prep Toolkit

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