Respuesta :

Notice the following pattern,

[tex]\begin{gathered} 8=2^3,4=2^2,2=2^1,0=2^0 \\ \frac{1}{2}=2^{-1},\frac{1}{4}=\frac{1}{2^2}=2^{-2},... \end{gathered}[/tex]

Therefore, the given sequence can be rewritten as shown below

[tex]8,4,2,1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},...=2^3,2^2,2^1,2^0,2^{-1},2^{-2},2^{-3},2^{-4},...[/tex]

Thus, the formula is

[tex]a_n=2^{4-n},n\ge1[/tex]

The n-th term of the sequence is given by 2^(4-n) (n=1,2,3,...)