Thread: method of differences

Find its sum :

\sum^{n}_{r=1}\frac{1}{(3n-2)(3n+1)(3n+4)}

After doing the partial fraction :

=\frac{1}{18}\sum^{n}_{r=1}(\frac{1}{3n-2}-\frac{1}{3n+1}+\frac{1}{3n+4})

Then from here , i tried putting f(r)=\frac{1}{3n-2} , 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:



You'll find

So,

Now apply the sum and see what happens.