Write the explicit formula that represents the geometric sequence -2, 8, -32, 128

Write the explicit formula that represents the geometric sequence -2, 8, -32, 128

$$\small{\text{$ \begin{array}{rcl} t_1 &=& -2\qquad r = \dfrac{8}{-2}=-4\\\\ t_n &=& t_1 \cdot r^{n-1}\\\\ t_n &=& (-2) \cdot (-4)^{n-1}\\\\ t_n &=& (-2) \cdot (-4)^n\cdot (-4)^{-1}\\\\ t_n &=& (-2)(-4)^{-1} \cdot (-4)^n\\\\ t_n &=& \frac{-2}{-4} \cdot (-4)^n\\\\ \mathbf{t_n} & \mathbf{=} & \mathbf{ \frac{1}{2} \cdot (-4)^n } \end{array} $}}$$

-2, 8, -32, 128

The "formula"  is given by:

(-1)ⁿ(2)^{2n - 1}  for n ≥ 1

I just wanted to muck around with this question and show how all the answers can be tied together :))

I would normally just have stopped on the first line :)

Write the explicit formula that represents the geometric sequence -2, 8, -32, 128

a=-2 r=-4

$$\\t_n=-2*(-4)^{n-1}\\\\ or\\\\ t_n=-2*(-4)^{n}(-4)^{-1}\\\\ t_n=\frac{-2}{-4}*(-4)^{n}\\\\ t_n=\frac{1}{2}*(-4)^{n}\\\\ t_n=\frac{1}{2}*(-1)^n*(4)^{n}\\\\ t_n=\frac{1}{2}*(-1)^n*(2^2)^{n}\\\\ t_n=\frac{1}{2}*(-1)^n*2^{2n}\\\\ t_n=(-1)^n*2^{2n-1}\\\\$$