If \(n\) is a positive integer, then \(((-3)^n)^{-4}+(3^{-n})^4+((-3)^{-n})^4=\)

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If \(n\) is a positive integer, then \(((-3)^n)^{-4}+(3^{-n})^4+((-3)^{-n})^4=\)

A. \(\left(\dfrac13\right)^{-4n}\)

B. \(\left(\dfrac13\right)^{-4n+1}\)

C. \(3^{-4n+1}\)

D. \(3^{-4n}\)

E. \(3^{-4n-1}\)

[spoiler]OA=C[/spoiler]

Source: e-GMAT
Source: — Problem Solving |

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M7MBA wrote:
Wed Jun 24, 2020 6:58 am
If \(n\) is a positive integer, then \(((-3)^n)^{-4}+(3^{-n})^4+((-3)^{-n})^4=\)

A. \(\left(\dfrac13\right)^{-4n}\)

B. \(\left(\dfrac13\right)^{-4n+1}\)

C. \(3^{-4n+1}\)

D. \(3^{-4n}\)

E. \(3^{-4n-1}\)

[spoiler]OA=C[/spoiler]

Source: e-GMAT
So, we have \(((-3)^n)^{-4}+(3^{-n})^4+((-3)^{-n})^4=?\)

Correct answer: C

Hope this helps!

-Jay
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=> \((-3)^{-4n}+3^{-4n}+(-3)^{-4n}\)

=> \((3)^{-4n}+3^{-4n}+(3)^{4n}\); note that since 4n is an even number, it is immaterial whether the base is positive or negative; the answer would always be positive.

=> \(3\cdot(3)^{-4n}\)

=> \((3)^{(-4n+1)}\)

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M7MBA wrote:
Wed Jun 24, 2020 6:58 am
If \(n\) is a positive integer, then \(((-3)^n)^{-4}+(3^{-n})^4+((-3)^{-n})^4=\)

A. \(\left(\dfrac13\right)^{-4n}\)

B. \(\left(\dfrac13\right)^{-4n+1}\)

C. \(3^{-4n+1}\)

D. \(3^{-4n}\)

E. \(3^{-4n-1}\)

[spoiler]OA=C[/spoiler]

Source: e-GMAT
Solution:

Simplifying each term, we have:

((-3)^n)^-4 = (-3)^(-4n) = 3^(-4n)

(3^-n)^4 = 3^(-4n)

((-3)^-n)^4 = (-3)^(-4n) = 3^(-4n)

Therefore, the sum is:

3 x 3^(-4n) = 3^(-4n + 1)

Answer: C

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