<s>[INST] Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that 
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$$
I tried taking $n=k$,
so it makes
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{k(k+3)}{4(k+1)(k+2)}.$$
Then proving the statement for $n=k+1$: 
$$\frac{k(k+3)}{4(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2) }.$$
 [/INST] You actually copied it over incorrectly; the general term should be $\frac{1}{n(n+1)(n+2)}$, not $\frac{1}{n(n+1)(n+3)}$. That in mind, try to see if you can follow this argument:
For each $n\geq 1$, let $S(n)$ denote the statement
$$
\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}.
$$
Base step ($n=1$): $S(1)$ says that $\frac{1}{1\cdot2\cdot3}=\frac{1\cdot4}{4\cdot2\cdot3}$, and this is correct since both sides equal $\frac{1}{6}$.
Inductive step ($S(k)\to S(k+1)$): For some fixed $k\geq 1$, assume the inductive hypothesis
$$
S(k) : \sum_{i=1}^k\frac{1}{i(i+1)(i+2)}=\frac{k(k+3)}{4(k+1)(k+2)}
$$
to be true. It remains to show that 
$$
S(k+1) : \sum_{i=1}^{k+1}\frac{1}{i(i+1)(i+2)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i(i+1)(i+2)} &= \sum_{i=1}^k\frac{1}{i(i+1)(i+2)}+\frac{1}{(k+1)(k+2)(k+3)}\\[1em]
&= \frac{k(k+3)}{4(k+1)(k+2)}+\frac{4}{4(k+1)(k+2)(k+3)}\tag{by $S(k)$}\\[1em]
&= \frac{k(k+3)(k+3)}{4(k+1)(k+2)(k+3)}+\frac{4}{4(k+1)(k+2)(k+3)}\\[1em]
&= \frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}\\[1em]
&= \frac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)}\\[1em]
&= \frac{(k+1)(k+4)}{4(k+2)(k+3)},
\end{align}
one arrives at the right-hand side of $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
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