If $n$ is a positive integer, then $\left(-2^{n}\right)^{-2}+\left(2^{-n}\right)^{2}=$

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

Explanation	Calculations	Help
$We simplify the first term by separating the factor - 1 and applying the rule for multiplying negative numbers an even number of times.$	$(- 2^{n})^{- 2} = (- 1)^{- 2} \times 2^{- 2 n} = 1 \times 2^{- 2 n} = 2^{- 2 n}$	Theory & Tactics Method Card PN2-B Click to view full details
$We simplify the second term by applying the power-of-a-power rule to the exponent.$	$(2^{- n})^{2} = 2^{- 2 n}$
$We add the two identical simplified terms to find the final expression.$	$2^{- 2 n} + 2^{- 2 n} = 2 \times 2^{- 2 n} = 2^{- 2 n + 1}$

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2^{- 2 n + 1}

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

Explanation	Calculations	Help
$We simplify the first term by separating the factor - 1 and applying the rule for multiplying negative numbers an even number of times.$	$(- 2^{n})^{- 2} = (- 1)^{- 2} \times 2^{- 2 n} = 1 \times 2^{- 2 n} = 2^{- 2 n}$	Theory & Tactics Method Card PN2-B Click to view full details
$We simplify the second term by applying the power-of-a-power rule to the exponent.$	$(2^{- n})^{2} = 2^{- 2 n}$
$We add the two identical simplified terms to find the final expression.$	$2^{- 2 n} + 2^{- 2 n} = 2 \times 2^{- 2 n} = 2^{- 2 n + 1}$

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2^{- 2 n + 1}