Ans: (a) $n(2n-3)$ (b) $64$
- $A(1) + A(2) + A(3) + \cdots + A(n)$
$\begin{array}{cl}
= & [4(1) – 5] + [4(2) – 5] + [4(3) – 5] + \cdots + [4(n) – 5 ] \\
= & -1 +3 + 7 + \cdots + 4n – 5 \\
= & \dfrac{[(-1) + (4n-5)]n}{2} \\
= & \dfrac{n(4n-6)}{2} \\
= & \dfrac{2n(2n-3)}{2} \\
= & n(2n-3)
\end{array}$ -
$\begin{array}{rcl}
\log (B(1)B(2)B(3) \cdots B(n)) & \le & 8000 \\
\log (10^{4(1)-5} \times 10^{4(2) -5} \times \cdots \times 10^{4n-5}) & \le & 8000 \\
\log (10^{[4(1)-5] + [4(2) -5] + \cdots + [4n-5]}) & \le & 8000 \\
\log (10^{n(2n-3)}) & \le & 8000 \\
n(2n-3) & \le & 8000 \\
2n^2 -3n – 8000 & \le & 0 \\
(2n+125)(n-64) & \le & 0
\end{array}$Therefore, $\dfrac{-125}{2} \le n \le 64$. Hence, the greatest value of $n$ is $64$.
