2020-M2-05

For $n=1$,

$\begin{array}{rclll}
\text{LS} & = & \dfrac{1}{(1)(1+1)(1+2)} & = & \dfrac{1}{6}
\end{array}$

$\begin{array}{rclll}
\text{RS} & = & \dfrac{(1)(1+3)}{4(1+1)(1+2)} & = & \dfrac{1}{6}
\end{array}$

$\therefore \text{LS} = \text{RS}$.

$\therefore P(1)$ is true.

Assume that $P(m)$ is true for some positive integers $m \ge 1$.

i.e. $\dsum_{k=1}^m \dfrac{1}{k(k+1)(k+2)} = \dfrac{m(m+3)}{4(m+1)(m+2)}$.

For $n=m+1$,

$\begin{array}{cl}
& \text{LS} \\
= & \dsum_{k=1}^{m+1} \dfrac{1}{k(k+1)(k+2)} \\
= & \dsum_{k=1}^m \dfrac{1}{k(k+1)(k+2)} +\dfrac{1}{(m+1)(m+1+1)(m+1+2)} \\
= & \dfrac{m(m+3)}{4(m+1)(m+2)} +\dfrac{1}{(m+1)(m+2)(m+3)} \\
= & \dfrac{m(m+3)(m+3)+4}{4(m+1)(m+2)(m+3)} \\
= & \dfrac{m(m^2+6m+9)+4}{4(m+1)(m+2)(m+3)} \\
= & \dfrac{m^3+6m^2+9m+4}{4(m+1)(m+2)(m+3)} \\
= & \dfrac{(m+1)(m^2+5m+4)}{4(m+1)(m+2)(m+3)} \\
= & \dfrac{(m+1)(m+1)(m+4)}{4(m+1)(m+2)(m+3)} \\
= & \dfrac{(m+1)(m+4)}{4(m+2)(m+3)} \\
= & \text{RS}
\end{array}$

Therefore, $P(m+1)$ is also true.

Therefore by the mathematical induction, $P(n)$ is true for all positive integers $n$.