Just try to use the common denominator:
$\displaystyle \frac13\left(1-\frac1{3n+1}\right)=
\frac13\left(\frac{3n+1}{3n+1}-\frac1{3n+1}\right)=
\frac13\cdot\frac{3n}{3n+1}=\frac{n}{3n+1}
$
which is exactly the RHS.
But the modification of the RHS is perhaps not so important.
Just try to use:
$\displaystyle \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)$
or (the same equation with n+1 instead of n
$\displaystyle \frac1{(3n+1)(3n+4)}=\frac13\left(\frac1{3n+1}-\frac1{3n+4}\right)$
in the inductive step of your proof.
EDIT: BTW the sums of this type are called
telescopic. With a little practice you can recognize easily at least the simple ones - and then you see the result immediately (that was the reason why I rewrote the RHS - since I saw that this expression is what I get from the telescopic sum.