Respuesta :

tonb
It sounds very difficult but I think it actually is quite simple.

For a sequence of ternary digits (ie., use symbols 0, 1 and 2), you have the freedom to choose the first digit as you like, but for all the other digits, you can only choose the 2 alternatives you have.

So for n digits the number of sequences is:

[tex]s(n) = 3 \cdot 2^{n-1} , n \in \mathbb{N} _{1} [/tex]

Rewriting this as a recurrence relation:

[tex]s(1)=3[/tex]
[tex]s(n+1) = 2s(n)[/tex]

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