How did you do this? It is wrong!

The correct identity is

$\displaystyle \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)} $

$\displaystyle \sum_{r=1}^{r=n} \frac{3r + 4}{ r(r+1)(r+2)} $

$\displaystyle = \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}$

$\displaystyle = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)$

$\displaystyle = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=1}^{r=n} \left(\frac{1}{r+1} - \frac{1}{r + 2}\right)$

$\displaystyle = \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=2}^{r= n+1} \left(\frac{1}{r} - \frac{1}{r + 1}\right)$

$\displaystyle = 2 - \frac2{n+1} + \frac12 - \frac1{n+2}$

$\displaystyle = 2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2}

$

EDIT: Oh yay , I got the right answer

And so on.... I might have made arithmetic mistakes... but I hope you get the general idea.