Solve 2*4^2(1-4^n-1)/1-4+4

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

Factor

$$\frac{4\left(2\times 4^{n}-5\right)}{3}$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$factor(2\times \frac{16\left(1-4^{n-1}\right)}{1-4}+4)$$

Subtract 4 from 1 to get -3.

$$factor(2\times \frac{16\left(1-4^{n-1}\right)}{-3}+4)$$

Express 2\times \frac{16\left(1-4^{n-1}\right)}{-3} as a single fraction.

$$factor(\frac{2\times 16\left(1-4^{n-1}\right)}{-3}+4)$$

Multiply 2 and 16 to get 32.

$$factor(\frac{32\left(1-4^{n-1}\right)}{-3}+4)$$

Use the distributive property to multiply 32 by 1-4^{n-1}.

$$factor(\frac{32-32\times 4^{n-1}}{-3}+4)$$

Factor out \frac{4}{3}.

$$\frac{4\left(-8\left(1-4^{n-1}\right)+3\right)}{3}$$

Evaluate

$$\frac{8\times 4^{n}-20}{3}$$