2012PP-I-19

$\begin{array}{rcl}
2~000~000+2~000~000(1-20\%) +\ldots +2~000~000(1-20\%)^{n-1} & \ge & 9~000~000 \\
2~000~000\times\dfrac{1-(0.8)^n}{1-0.8} & \ge & 9~000~000 \\
1-(0.8)^n & \ge & 0.9 \\
(0.8)^n & \le & 0.1 \\
n\log 0.8 & \le & \log 0.1 \\
n & \ge & \dfrac{\log 0.8}{\log 0.1} \\
n & \ge & 10.318~851~16
\end{array}$

Therefore, the least number of years needed is $11$.

$\begin{array}{rcl}
2~000~000+2~000~000(1-20\%) + \ldots +2~000~000(1-20\%)^{m-1} & \ge & 10~000~000 \\
2~000~000\times\dfrac{1-(0.8)^m}{1-0.8} & \ge & 10~000~000 \\
1-(0.8)^m & \ge & 1 \\
(0.8)^m & \le & 0
\end{array}$

It is impossible to find a real number $m$ such that $(0.8)^m \le 0$.

Therefore, the total revenue made by the firm will not exceed $\$10~000~000$.

$\begin{array}{cl}
& \mbox{Total revenue} – \mbox{Total investment} \\
= & \dfrac{2~000~000(1-(0.8)^k)}{1-0.8}-\dfrac{4~000~000(1-(0.64)^k)}{1-0.64} \\
= & 10^7(1-(0.8)^k) – \dfrac{10^8}{9}(1-(0.64)^k) \\
= & \dfrac{10^7}{9}[(9(1-(0.8)^k)-10(1-(0.64)^k)] \\
= & \dfrac{10^7}{9}[10(0.64)^k – 9(0.8)^k-1] \\
= & \dfrac{10^7}{9}[10(0.8)^2k-9(0.8)^l-1] \\
= & \dfrac{10^7}{9}(10(0.8)^k+1)((0.8)^k-1)
\end{array}$

Note that $(0.8)^k-1<0$ and $10(0.8)^k+1>0$ for any value of $k$.

Hence, we have $\dfrac{10^7}{9}(10(0.8)^k+1)((0.8)^k-1)<0$.

Therefore, total revenue $-$ total investment $<0$. Therefore I disagree the claim.