Formula for $\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$?

I was trying to find a formula for the series

$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} =? $$

I tried to break this into partial fractions...To see if I could telescope this series.. The partial fraction went like this $$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2k+4}$$

But the terms are so random that they hardly cancel.... I also tried partial sums but couldn't make much headway....Any help to solve this would be appreciated

$$\dfrac2{k(k+1)(k+2)}=\dfrac{k+2-k}{k(k+1)(k+2)}=\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}=f(k)-f(k+1)$$

where $f(m)=\dfrac1{m(m+1)}$

HINT$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} = \sum_{k=1}^n \frac{1}{2} \left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)} \right)$$ Can you take it from here?