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

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

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

simplify

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

4

Solution steps

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

Factor

2^{n + 2} + 2^{n} : 5 \cdot 2^{n}

= \frac{5 \cdot 2 ^{n}}{2 ^{n - 2} + 2 ^{n}}

Factor

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

= \frac{5 \cdot 2 ^{n}}{( \frac{1}{4} + 1 ) \cdot 2 ^{n}}

Cancel the common factor:

2^{n}

= \frac{5}{\frac{1}{4} + 1}

Join

\frac{1}{4} + 1 : \frac{5}{4}

= \frac{5}{\frac{5}{4}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{5 \cdot 4}{5}

Multiply the numbers:

5 \cdot 4 = 20

= \frac{20}{5}

Divide the numbers:

\frac{20}{5} = 4

= 4

a^{2} - a - (x^{2} + 7 x - 12)

12 x - 4 x^{5} x^{2}

x^{2} + x^{1}

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

\frac{( 5 \cdot 3 ^{n} - 9 \cdot 3 ^{n - 2} )}{( 3 ^{n} - 3 ^{n - 1} )}

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