Sum $‎‎\sum_{n=0}^{5‎}{n}{(n+1)}‎‎$ Since this series seems to be 0,2,6,12,20 and 30 there isn't a common difference that is identical between each term - it in itself seems to increase by 2 each time. 

I can't use $S_n=\frac{n[2a+(n-1)d]}{2}$ since I don't know what to do about the common difference.

Answer is supposed to be 112.

$$‎‎\sum_{n=0}^{5‎}{n}{(n+1)}‎‎$$
You could use the formula
$$\sum_{n=0}^N n^2=\frac{N(N+1)(2N+1)}{6}$$
together with the AP formula to get
$$\sum_{n=0}^5n(n+1)=\sum_{n=0}^5n^2+\sum_{n=0}^5n=\frac{5\times6\times11}{6}+\frac{5\times6}{2}=55+15=70\ .$$
On the other hand, since you have just worked out all the terms and there are not many of them, you may as well just add them up...

The answer would be $112$ if the upper limit of summation was $6$ instead of $5$.
Another way to do this involves converting the sum to a sum of a telescoping series and a trivial series.  Consider

$$(n+1)^3-n^3 = 3 n^2+3 n+1 = 3 n (n+1)+1 \implies n (n+1) = \frac13 [(n+1)^3-n^3] - \frac13 $$

Thus,

$$\sum_{n=1}^{N} n (n+1) = \frac13 \sum_{n=1}^{N} [(n+1)^3-n^3] - \frac13 \sum_{n=1}^{N} 1 = \frac13 [(N+1)^3 - 1] - \frac13 N$$

You can simplify this to get

$$\sum_{n=1}^{N} n (n+1) = \frac13 (N^3 + 3 N^2 +2 N) = \frac13 N (N+1)(N+2)$$

Plug in $N=5$ to get your answer.