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 $2+6+12+20+30+42+56+72+\dots$
Yes there is. Ever wonder why this is called quadratic sequence? Quadratic refers to squares right? This is just constant difference of difference. So where's the connection? Well as it turns out, all terms of a quadratic sequence are expressible by a quadratic polynomial. What do I  mean? Consider this

$$
t_n = n+n^2
$$

Subsituiting $n=1,2,3,\cdots$ generates your terms. By the way, $202$ doesn't occur in this sequence, the 13th term is $182$ and the $14th$ term is $210$. I am assuming it was supposed to be $210$. 

So we need to find

$$
\sum_{i=1}^{n}i+i^2 = \sum_{i=1}^{n}i+\sum_{i=1}^{n}i^2
$$

where $n=14$. There are well known formulas for $\sum_{i=1}^{n}i$ and for $\sum_{i=1}^{n}i^2$. Substituting them, we get,

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

where $n=14$. Thus our sum is $1120$.
I have a sum formula for quadratic equation.

$$s=\frac{n}{6} (3d(n-1)+(n-1)(n-2)c) +an $$

Where $n$ is the number of terms, $d$ is the first difference, $c$ is the constant difference or difference of difference, $a$ is first term. For calculation of sum first put value of $a$ $d$ $c$ then you get a formula like $an^2+bn+c$.