Prove that a sequence whose second difference is a nonzero constant is quadratic. For example, if {$a_0, a_1, a_2, a_3, ...$} is the sequence, 
the first difference is {$a_1-a_0, a_2-a_1, a_3-a_2, ...$}, 
and the second difference is {$(a_2-a_1)-(a_1-a_0), (a_3-a_2)-(a_2-a_1), ...$}.
I think that using facts from up to Calculus, perhaps derivatives, should be enough. I find myself going in circles and don't know how to approach this.
 A: Let $d$ denote the constant second difference. Moreover, let
\begin{align*}
c\equiv&\,a_1-a_0-\frac{d}{2},\\
\end{align*}

Claim: $a_n=(d/2)n^2+cn+a_0$ for all $n\in\{0,1,2\ldots\}$.

Proof: The claim is obviously true for $n=0$. For $n=1$, $$\frac{d}{2}\times n^2+cn+a_0=\frac{d}{2}+\left(a_1-a_0-\frac{d}{2}\right)+a_0=a_1.$$
Proceed by induction: suppose that the claim is true for $0,1,\ldots,n$ for some integer $n\geq1$. The task is to prove that it is true for $n+1$. Now: 
\begin{align*}
d=&\,(a_{n+1}-a_n)-(a_n-a_{n-1})=a_{n+1}-2a_n+a_{n-1}\\
=&\,a_{n+1}-2\left[\frac{d}{2}\times n^2+cn+a_0\right]+\left[\frac{d}{2}\times (n-1)^2+c(n-1)+a_0\right],
\end{align*}
where the first equality comes from the definition of $d$, and the third one is due to the induction hypothesis. Now, rearrange for $a_{n+1}$:
\begin{align*}
a_{n+1}=&\,d+2\left[\frac{d}{2}\times n^2+cn+a_0\right]-\left[\frac{d}{2}\times (n-1)^2+c(n-1)+a_0\right]\\
=&\,d+dn^2+2cn+2a_0-\frac{d}{2}\times(n^2-2n+1)-c(n-1)-a_0\\
=&\,\underbrace{d}_{\spadesuit}+\underbrace{dn^2}_{\heartsuit}+\underbrace{2cn}_{\clubsuit}+\underbrace{2a_0}_{\diamondsuit}-\underbrace{\frac{d}{2}\times n^2}_{\heartsuit}+\underbrace{dn}_{\star}-\underbrace{\frac{d}{2}}_{\spadesuit}-\underbrace{cn}_{\clubsuit}+\underbrace{c}_{\clubsuit}-\underbrace{a_0}_{\diamondsuit}\\
=&\,\underbrace{\frac{d}{2}}_{\spadesuit}+\underbrace{\frac{d}{2}\times n^2}_{\heartsuit}+\underbrace{c(n+1)}_{\clubsuit}+\underbrace{a_0}_{\diamondsuit}+\underbrace{dn}_{\star}\\
=&\,\frac{d}{2}\times n^2+dn+\frac{d}{2}+c(n+1)+a_0\\
=&\,\frac{d}{2}\times(n+1)^2+c(n+1)+a_0.
\end{align*}
The proof is complete. $\quad\blacksquare$
