Is there a general formula for the sum of a quadratic sequence? I tried Googling "formula for sum of quadratic sequence", which did not give me anything useful. I just want an explicit formula for figuring out a sum for a quadratic sequence. For example, how would you figure out the sum of $2+6+12+20+\dots+210$? Can someone please help? Thanks
For those of you who do not know, a quadratic sequence is a sequence where the differences of the differences between the terms are constant. Let's use $2+6+12+20+\dots$ as an example. The differences between the terms are $4$, $6$, $8$, etc. The difference between the differences of the terms is $2$. So the sequence will continue like $2+6+12+20+30+42+56+72+\dots$

In this particular case, 
$$\begin{align}
&2+6+12+20+\cdots+210\\\\
&=2(1+3+6+10+\cdots+105)\\\\
&=2\left[\binom 22 +\binom 32 +\binom 42+\binom 52+\cdots \binom {15}2\right]\\
&=2\sum_{r=2}^{15}\binom r2\\
&=2\binom {16}3\\
&=\color{red}{1120}\qquad\blacksquare
\end{align}$$

This is the sum of triangular numbers (where the difference of the difference is constant) and the result is a pyramidal number (all scaled by 2). The summation can be shown as 
$$\begin{align}
&2\cdot (1)+\\
&2\cdot (1+2)+\\
&2\cdot (1+2+3)+\\
&2\cdot (1+2+3+4)+\\
&\quad \vdots\qquad\qquad\qquad\ddots\\
&2\cdot (1+2+3+\cdots+15)
\end{align}$$
If we tried to sum a series where the difference of the difference of the difference is constant, i.e. sum of pyramidal numbers, the result would be a pentatope number. And so on...
An example of the summation of pyramidal numbers, extending from the original question, would be
$$2+8+20+40+\cdots+910 =2 \sum_{r=1}^{15} \binom r3=2 \binom {16}4=1820\\$$
