<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] Given that $$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$$
1. Substituting $\color{blue}{n=1}$ in the given equality, we get $$\frac{1}{1\times 2\times 3}=\frac{(1)(1+3)}{4(1+1)(1+2)}$$ $$\implies \frac{1}{6}=\frac{4}{24}\implies \frac{1}{6}=\frac{1}{6}$$ Hence, it holds for $n=1$


*Assuming that it holds for $\color{blue}{n=k}$ then we have $$\color{blue}{\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}}$$

*Now, substituting $\color{blue}{n=k+1}$ in the given equality, we have $$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}$$ 
$$\implies \frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{k(k+1)(k+2)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}-\frac{1}{(k+1)(k+2)(k+3)}$$  $$=\frac{1}{(k+2)(k+3)}\left(\frac{(k+1)^2(k+4)-4}{4(k+1)}\right)$$ $$=\frac{k^3+2k^2+k+4k^2+8k+4-4}{4(k+1)(k+2)(k+3)}$$ $$=\frac{k(k^2+6k+9)}{4(k+1)(k+2)(k+3)}$$ $$=\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}$$ $$=\frac{k(k+3)}{4(k+1)(k+2)}$$ Which is true from (2) thus it also holds for $n=k+1$
Hence, it is clear from (1), (2) & (3), the given equality holds for all $\color{blue}{n\geq 1}$ 
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