Let $a_n$ and $b_n$ be the number of A's and B's respectively in the $\mathrm{nth}$ word and $f_n$ the total number of letters in the $\mathrm{nth}\mathrm{word}$. Note that in each word there is an A for every B in the previous word so

${\displaystyle a_n=b_{n-1}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1).}$

The number of B's is given by the number of B's in the previous word plus the number of A's in the previous word and so

${\displaystyle b_{n+1}=a_n+b_n\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)}$

. Putting these two expressions together and substituting for $a_n$ in (2) we get

${\displaystyle b_{n+1}=b_{n-1}+b_n}$

so the sequence $b_n$ is a Fibonacci sequence and the pattern will continue. Similarly substituting for $b_n$ in (2) we get

${\displaystyle a_{n+2}=a_n+a_{n+1}}$

so the $a_n$ forn a Fibonaci sequence and the pattern will continue.

Because the two sequences of numbers are the same apart from the shift of one place the total number of letters is also a Fibonacci sequence and the pattern will continue.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{a_n}& =b_{n-1}\\ \multicolumn{1}{c}{f_n}& =a_n+b_n=b_{n-1}+b_n=b_{n+1}\\ \multicolumn{1}{c}{f_{n-1}}& =a_{n-1}+b_{n-1}=a_{n-1}+a_n=a_{n+1}\\ \multicolumn{1}{c}{f_{n+1}}& =a_{n+1}+b_{n+1}=f_{n-1}+f_n.\end{array}}$