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

The solution to ((3^{n+1}+3^n))/((3^{n+3)-3^{n+1})} is 1/6

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

simplify

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

\frac{3 ^{n + 1} + 3 ^{n}}{3 ^{n + 3} - 3 ^{n + 1}}

Factor

3^{n + 1} + 3^{n} : 4 \cdot 3^{n}

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

Factor

3^{n + 3} - 3^{n + 1} : 8 \cdot 3^{1 + n}

= \frac{4 \cdot 3 ^{n}}{8 \cdot 3 ^{1 + n}}

Cancel

\frac{4 \cdot 3 ^{n}}{8 \cdot 3 ^{1 + n}} : \frac{1}{2 \cdot 3}

= \frac{1}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1}{6}

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What is ((3^{n+1}+3^n))/((3^{n+3)-3^{n+1})} ?
The solution to ((3^{n+1}+3^n))/((3^{n+3)-3^{n+1})} is 1/6