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 $n=k+1$: 
$$\frac{k(k+3)}{4(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2) }.$$

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*}
