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

The solution to ((2^{n+1}+2^n))/((2^{n+4)-2^{n+1})} is 3/14

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

simplify

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

\frac{3}{14}

Decimal Notation

0.21428 \dots

Solution steps

\frac{2 ^{n + 1} + 2 ^{n}}{2 ^{n + 4} - 2 ^{n + 1}}

Factor

2^{n + 1} + 2^{n} : 3 \cdot 2^{n}

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

Factor

2^{n + 4} - 2^{n + 1} : 7 \cdot 2^{1 + n}

= \frac{3 \cdot 2 ^{n}}{7 \cdot 2 ^{1 + n}}

Cancel

\frac{3 \cdot 2 ^{n}}{7 \cdot 2 ^{1 + n}} : \frac{3}{7 \cdot 2}

= \frac{3}{7 \cdot 2}

Multiply the numbers:

7 \cdot 2 = 14

= \frac{3}{14}

8 x^{3} - (2 x - y)^{3}

x^{29} - x^{24} + x^{13} - 1

(\frac{x y + 1}{2})^{3}

sec^{2} (4 5^{\circ}) x

\frac{\frac{s}{n}}{161906}

What is ((2^{n+1}+2^n))/((2^{n+4)-2^{n+1})} ?
The solution to ((2^{n+1}+2^n))/((2^{n+4)-2^{n+1})} is 3/14