What is (3^{n+4}-3(3)^{n-1})/(3(3^{n+2))} ?

The solution to (3^{n+4}-3(3)^{n-1})/(3(3^{n+2))} is 80/27

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simplify (3^{n+4}-3(3)^{n-1})/(3(3^{n+2))}

simplify

\frac{3 ^{n + 4} - 3 ( 3 ) ^{n - 1}}{3 ( 3 ^{n + 2} )}

\frac{80}{27}

Decimal Notation

2.96296 \dots

Solution steps

\frac{3 ^{n + 4} - 3 \cdot 3 ^{n - 1}}{3 ( 3 ^{n + 2} )}

Remove parentheses:

(a) = a

= \frac{3 ^{n + 4} - 3 \cdot 3 ^{n - 1}}{3 \cdot 3 ^{n + 2}}

3^{n + 4} - 3 \cdot 3^{n - 1} = 3^{n + 4} - 3^{n}

= \frac{3 ^{n + 4} - 3 ^{n}}{3 \cdot 3 ^{n + 2}}

3 \cdot 3^{n + 2} = 3^{n + 3}

= \frac{3 ^{n + 4} - 3 ^{n}}{3 ^{n + 3}}

Factor

3^{n + 4} - 3^{n} : 80 \cdot 3^{n}

= \frac{80 \cdot 3 ^{n}}{3 ^{n + 3}}

Cancel

\frac{80 \cdot 3 ^{n}}{3 ^{n + 3}} : \frac{80}{3 ^{3}}

= \frac{80}{3 ^{3}}

3^{3} = 27

= \frac{80}{27}

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What is (3^{n+4}-3(3)^{n-1})/(3(3^{n+2))} ?
The solution to (3^{n+4}-3(3)^{n-1})/(3(3^{n+2))} is 80/27