Solution:
Solution
$$B=\left[ \begin{matrix} 3 & 4 \\ 2 & 3 \end{matrix} \right] $$ $$;$$ $$C=\left[ \begin{matrix} 3 & -4 \\ -2 & 3 \end{matrix} \right] $$$$\Rightarrow BC=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] $$ $$\Rightarrow BC=I$$
$$\therefore ABC=A.I=A$$
$$A\left( BC\right)^2=AI^2=A$$
$$\therefore tr\left( A\right) +tr\left( \dfrac{ABC}{2}\right) +tr. \left( \dfrac{A\left( BC\right) ^2}{4}\right) +tr. \left( \dfrac{A\left( BC\right)^3}{8}\right)+........$$
$$=3+\dfrac{3}{2}+\dfrac{3}{4}+\dfrac{3}{8}+......+A$$
$$=3\left( \dfrac{1}{1-\dfrac{1}{2}}\right) $$
$$=\dfrac{3}{\dfrac{1}{2}}$$
$$=6$$
Hence, the answer is $$6.$$