Evaluate a=(32^n-42^(n-2))/(2^n-2^(n-1))

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

Step 1

Simplify the numerator.

Multiply $- 4 \cdot 2^{n - 2}$ .

Factor out negative.

$a = \frac{3 \cdot 2^{n} - (4 \cdot 2^{n - 2})}{2^{n} - 2^{n - 1}}$

Rewrite $4$ as $2^{2}$ .

$a = \frac{3 \cdot 2^{n} - (2^{2} \cdot 2^{n - 2})}{2^{n} - 2^{n - 1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Subtract $2$ from $2$ .

$a = \frac{3 \cdot 2^{n} - 2^{n + 0}}{2^{n} - 2^{n - 1}}$

Add $n$ and $0$ .

$a = \frac{3 \cdot 2^{n} - 2^{n}}{2^{n} - 2^{n - 1}}$

Subtract $2^{n}$ from $3 \cdot 2^{n}$ .

$a = \frac{2 \cdot 2^{n}}{2^{n} - 2^{n - 1}}$

Step 2

Multiply $2$ by $2^{n}$ .

Raise $2$ to the power of $1$ .

$a = \frac{2^{1} \cdot 2^{n}}{2^{n} - 2^{n - 1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$a = \frac{2^{1 + n}}{2^{n} - 2^{n - 1}}$

Evaluate a=(3*2^n-4*2^(n-2))/(2^n-2^(n-1))