
method of differences
Find its sum :
\sum^{n}_{r=1}\frac{1}{(3n2)(3n+1)(3n+4)}
After doing the partial fraction :
=\frac{1}{18}\sum^{n}_{r=1}(\frac{1}{3n2}\frac{1}{3n+1}+\frac{1}{3n+4})
Then from here , i tried putting f(r)=\frac{1}{3n2} , and f(r+1)=\frac{1}{3n+1} .
Putting these functions back into the summation doesn't work because there is an extra term there .
What can i do from here ?

You better split the fraction this way:
$\displaystyle \frac{1}{(3n2)(3n+1)(3n+4)}=\frac{A}{(3n2)(3n+1)}+\frac{B}{(3n+1)(3n+4)}$
You'll find $\displaystyle A=\frac{1}{6}, \ B=\frac{1}{6}$
So, $\displaystyle \frac{1}{(3n2)(3n+1)(3n+4)}=\frac{1}{6}\left(\frac{1}{(3n2)(3n+1)}\frac{1}{(3n+1)(3n+4)}\right)$
Now apply the sum and see what happens.