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 $n=k+1$:
$$\frac{k(k+3)}{4(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2) }.$$
On
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 $\color{blue}{n\geq 1}$
On
Actually, mathematical induction is not necessary for this formula. A direct calculation with a telescoping decomposition into partial fractions will show where the formula comes from.
Indeed one checks that $$\frac1{k(k+1)(k+2)}=\frac12\frac1k-\frac1{k+1}+\frac12\frac1{k+2}.$$ Thus the sum is: \begin{align*}&\Bigl(\frac12-\frac12+\frac12\frac13\bigr)+\Bigl(\frac12\frac12-\frac13+\frac12\frac14\Bigr)+\Bigl(\frac12\frac13-\frac14+\frac12\frac15\bigr)+\dotsm\dotsm\\&\color{red}+\frac1{2(n-2)}-\frac1{n-1}+\frac1{2n}\color{red}+\frac1{2(n-1)}-\frac1n+\frac1{2(n+1)}\color{red}+\frac1{2n}- \frac1{n+1} +\frac1{2(n+2)}\\ &=\frac14+\frac12\Bigl(\frac1{n+2}-\frac1{}\Bigr)=\frac{(n+1)(n+2)-2}{4(n+1(n+2)} = \frac{n(n+3)}{4(n+1(n+2)}. \end{align*}
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 $n\geq 1$. $\blacksquare$