summation - Calculating sums

My maths teacher showed me something on how to calculate sums. Let's take an example:

$$\sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$$

This was an easy one, but I just can't understan how to solve such sums:

$$\sum_{k=1}^n (k-1)k(k+1)\tag{example 1}$$

$$\sum_{k=1}^n \frac{1}{(3n-2)(3n+1)}\tag{example 2}$$

Could anybody help me, please?

I want to understand the idea of solving sums like these, so please, do not be very specific, but help me giving these and maybe some other examples.

Answer

HINT:

The second example is orthogonal to the first, hence a different answer

$$\frac3{(3n+1)(3n-2)}=\frac{(3n+1)-(3n-2)}{(3n+1)(3n-2)}=\frac1{3n-2}-\frac1{3n+1}$$

Set a few values of $n=1,2,3,\cdots,n-2,n-1,n$ to recognize the Telescoping Series