What is (2^{2n+1}+4^{n-1}+16^{n/2})/(2^{2n-2)} ?

The solution to (2^{2n+1}+4^{n-1}+16^{n/2})/(2^{2n-2)} is (2^{2n+1}+4^{n-1}+4^n)/(2^{2n-2)}

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simplify (2^{2n+1}+4^{n-1}+16^{n/2})/(2^{2n-2)}

simplify

\frac{2 ^{2 n + 1} + 4 ^{n - 1} + 1 6 ^{\frac{n}{2}}}{2 ^{2 n - 2}}

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

Solution steps

\frac{2 ^{2 n + 1} + 4 ^{n - 1} + 1 6 ^{\frac{n}{2}}}{2 ^{2 n - 2}}

Factor the number:

16 = 4^{2}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

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

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

Simplify

2 \cdot \frac{n}{2} : n

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

\frac{9 y ^{2}}{y - 2} - \frac{y ^{2} + 16 y}{y - 2}

\frac{3 x ^{2} - 11 x + 6}{9 x ^{2} - 12 x + 4}

(- 2 x^{3}) (- 5 x) (- 3 x^{2})

2 x^{3} - x^{2}

\frac{64}{24}

What is (2^{2n+1}+4^{n-1}+16^{n/2})/(2^{2n-2)} ?
The solution to (2^{2n+1}+4^{n-1}+16^{n/2})/(2^{2n-2)} is (2^{2n+1}+4^{n-1}+4^n)/(2^{2n-2)}