Grade 11 math again

Find the 10th term of the sequence 2, 8, 32, 128....

The general form of a geometric sequence is

${\displaystyle a,\ ar,\ ar^{2},\ ar^{3},\ ar^{4},\ \ldots }$

The common ratio r is  $\frac{8}{2}=\frac{32}{8}=\frac{128}{32}=4$

a is a scale factor, equal to the sequence's start value = 2

The n-th term of a geometric sequence with initial value a and common ratio r is given by

$\begin{array}{rcll} a_{n} &=& a\,r^{n-1} \quad &| \quad a=2 \quad r = 4 \\ a_{n} &=& 2\cdot 4^{n-1} \\ \end{array}$

The 10th term is
$\begin{array}{|rcll|} \hline a_{10} &=& 2\cdot 4^{10-1} \\ a_{10} &=& 2\cdot 4^{9} \\ a_{10} &=& 2\cdot 262144 \\ \mathbf{ a_{10}} & \mathbf{=} & \mathbf{524288} \\ \hline \end{array}$

Is that 328  supposed to be 128?

I think this should be    2, 8, 32 , 128.....

The 10h term is given by  :    2^{2n - 1}   =  2^{2(10) - 1}  = 2¹⁹  = 524,298

Slight correction CPhill: a(n) =F x R^(N-1) =2 x 4^9=524,288.

THX for the correction, Guest.....