# method of differences

• September 3rd 2009, 08:11 AM
thereddevils
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 ?
• September 3rd 2009, 08:27 AM
red_dog
You better split the fraction this way:

$\frac{1}{(3n-2)(3n+1)(3n+4)}=\frac{A}{(3n-2)(3n+1)}+\frac{B}{(3n+1)(3n+4)}$

You'll find $A=\frac{1}{6}, \ B=-\frac{1}{6}$

So, $\frac{1}{(3n-2)(3n+1)(3n+4)}=\frac{1}{6}\left(\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}\right)$

Now apply the sum and see what happens.