I2.1 Given that $P=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{99 \times 100}$, find the value of $P$.
I2.2 Given that $99 Q=P \times\left(1+\frac{99}{100}+\frac{99^{2}}{100^{2}}+\frac{99^{3}}{100^{3}}+\ldots\right)$, find the value of $Q$.
I2.3 Given that $x$ and $R$ are real numbers and $\frac{2 x^{2}+2 R x+R}{4 x^{2}+6 x+3} \leq Q$ for all $x$, find the maximum value of $R$.

I2.4 Given that $S=\log _{144} \sqrt[R]{2}+\log _{144} \sqrt[R]{R}$, find the value of $S$.
$$
\begin{array}{l}
P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{99}-\frac{1}{100} \\
=1-\frac{1}{100}=\frac{99}{100} \\
99 Q=\frac{99}{100} \times\left(1+\frac{99}{100}+\frac{99^{2}}{100^{2}}+\frac{99^{3}}{100^{3}}+\ldots\right) \\
\quad=\frac{99}{100} \times \frac{1}{1-\frac{99}{100}}=99 \\
Q=1 \\
4 x^{2}+6 x+3=(2 x+1.5)^{2}+0.75>0 \\
\frac{2 x^{2}+2 R x+R}{4 x^{2}+6 x+3} \leq 1 \\
2 x^{2}+2 R x+R \leq 4 x^{2}+6 x+3 \\
2 x^{2}+2(3-R) x+3-R \geq 0 \\
\Delta \leq 0 \\
(3-R)^{2}-2(3-R) \leq 0 \\
(3-R)(1-R) \leq 0 \\
1 \leq R \leq 3
\end{array}
$$

The maximum value of $R=3$
$$
S=\frac{\frac{1}{3} \log 2}{\log 144}+\frac{\frac{1}{6} \log 3}{\log 144}=\frac{2 \log 2+\log 3}{6 \log 144}=\frac{\log 12}{6 \log 12^{2}}=\frac{\log 12}{12 \log 12}=\frac{1}{12}
$$

Method 2
$$
\begin{aligned}
S & =\log _{144} \sqrt[R]{2}+\log _{144} \sqrt[2 R]{R} \\
& =\log _{144}(\sqrt[3]{2} \cdot \sqrt[6]{3}) \\
& =\log _{144}(\sqrt[6]{12}) \\
& =\log _{144}(\sqrt[12]{144})=\frac{1}{12}
\end{aligned}
$$