Why does $\sum\limits_{i=1}^n i^2 = An^3+Bn^2+Cn + D$? I've got this question because of this video (around 3:15). I wonder how setting up a system of $3$ equations will help him solve this problem. I'm thinking I might not understand this because I've never really used a system of equations before to solve real life problems.
 A: This is another way you might gain an intuition for why such a form works.
So you want a closed form for the sum $\displaystyle \sum^{n}_{i = 1} i^2$
First consider all numbers of the form $\displaystyle k^3 - (k - 1)^3$, and sum this series from $1$ to $n$, which gives us  $\displaystyle \sum^{n}_{1} k^3 - (k - 1)^3$. Now notice that all the entries of this series cancels out except for the $n^3$ and $0^3$ terms, but since $0^3 = 0$, we have that $\displaystyle \sum^{n}_{1} k^3 - (k - 1)^3 = n^3$ (If you want to see this clearly, just write out the whole sum and then see the cancellation occuring!)
But, via simple algebraic manipulation $\displaystyle k^3 - (k - 1)^3 = k^3 - (k^3 - 3k^2 + 3k - 1) = 3k^2 - 3k + 1$.
So, $\displaystyle \sum^{n}_{1} k^3 - (k - 1)^3 = n^3 = \sum^{n}_{1}(3k^2 - 3k + 1) = \sum^{n}_{1}3k^2 - \sum^{n}_{1}3k + \sum^{n}_{1} 1$.
So we have that $\displaystyle n^3 = \sum^{n}_{1}3k^2 - \sum^{n}_{1}3k + \sum^{n}_{1} 1 \Rightarrow n^3 + \sum^{n}_{1}3k - \sum^{n}_{1} 1 = 3\sum^{n}_{1}k^2$.
Now, you should remind yourself (or prove via a very similar method to what I have used so far) that $\displaystyle \sum^{n}_{1}k = \dfrac{n(n + 1)}{2}$, and that $\displaystyle \sum^{n}_{1} 1$ is simply $n$ (which is easy enough to see since you're summing $n$ $1$s).
So the maximum degree of the polynomial $\displaystyle n^3 + \sum^{n}_{1}3k - \sum^{n}_{1} 1 $ can only be $3$ at most since all other terms have degree less than $3$. This means the closed form for $\displaystyle 3\sum^{n}_{1}k^2$ is also a polynomial (of $n$) of degree 3 at most, and since that $3$ in front of the sum is just a constant, you can divide it out.
So $\displaystyle \sum^{n}_{1}k^2 = An^3 + Bn^2 + Cn + D$
You can extend this method to find the closed sum of the sum of series of any powers, and infact, via induction, you can show that $\displaystyle \sum^{n}_{i = 1} i^k = An^{k + 1} + Bn^{k} + Cn^{k - 1} + ...$.
A: You can prove this the following way:
Show first that $k^2 = A\left(k^3-(k-1)^3 \right)+B\left(k^2-(k-1)^ 2\right)+C\left(k-(k-1) \right)$.
Then sum from $k=1$ to $k=n$ and use the fact that the RHS is telescopic.
This shows in general that 
$$\sum\limits_{i=1}^n i^k = An^{k+1}+Bn^k+Cn^{k-1} + ...$$
If you know that this formula holds, you can find $A,B,C,...$ by finding them in the first equation.
Aternately, writing the equations for $n=1, n=2, n=3, ... ,n=k+1$ you get a system.
Added: if you know linear algebra: The polynomials $X^3-(X-1)^3 \,;\, X^2-(X-1)^2\,;\, X-(X-1)$ have pairwise distinct degrees, thus they are linearly independent. Therefore, they are a basis for the space of polynomials of degree at most two. Hence $X^2$ can be written as a linear combination of these polynomials.
If you don't know linear algebra
Note that by opening the brackets we have
$$A(k^3-(k-1)^3)+B(k^2-(k-1)^2)+C(k-(k-1))= A( 3k^2-3k+1)+B(2k-1)+C \\
=k^2 ( 3A )+ k(-3A +2B)+(A-B+C)$$
Since we want to get exactly $k^2$ we want
$$3A=1 \\
2B=3A=1 \\
C=B-A \,.$$
This shows that, unless I made a mistake by typing too fast, that $A=\frac{1}{3}, B=\frac{1}{2}, C=\frac{1}{6}$.
We proved that they exists by finding them.
A: Let $\delta$ the backward difference operator:
$$\delta p(x)=p(x)-p(x-1).$$
If $p$ is a polynomial with degree $n$, then $\delta p$ is a polynomial with degree $n-1$. 
Since $\delta$ is invertible, for any $k\in\mathbb{N}$ we have that
$$ \sum_{n=1}^{N}n^k $$
is a polynomial in $N$ with degree $k+1$. In our case,
$$ \sum_{n=1}^{N} n^2 = A N^3 + B N^2 + C N + D  \tag{1} $$
and the values of $A,B,C,D$ can be found by imposing that $(1)$ holds for $N=1,2,3,4.$ We get:
$$\left\{\begin{array}{rcl}A+B+C+D &=& 1,\\ 8A+4B+2C+D &=& 5,\\ 27A+9B+3C+D&=&14, \\ 64A+16B+4C+D &=& 30.\end{array}\right.\tag{2} $$
Notice that we need four equations since the RHS of $(1)$ has four unknown coefficients. 
Now $(2)$ can be solved with many techniques, its solution is:
$$ A = \frac{1}{3},\quad B=\frac{1}{2},\quad C=\frac{1}{6},\quad D=0.\tag{3}$$
There is a solution (and only one) for any system like $(2)$ since the Vandermonde matrix is invertible.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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One nice way to check this is the Stolz-Cesaro Theorem. Lets
$\ds{\,{\rm S}_{k}\pars{n} \equiv \sum_{i\ =\ 0}^{n}i^{k}}$. Note that, with
$\ds{s = 1,2,3,\ldots}$, 
\begin{align}
\lim_{n\ \to\ \infty}\ {\,{\rm S}_{k}\pars{n} \over n^{s}}
&=
\lim_{n\ \to\ \infty}\ {\,{\rm S}_{k}\pars{n + 1} - \,{\rm S}_{k}\pars{n}
 \over \pars{n + 1}^{s} - n^{s}}
=\lim_{N\ \to\ \infty}\ {\pars{n + 1}^{k} \over sn^{s - 1}}
={1 \over s}\lim_{n\ \to\ \infty}\ n^{k - s + 1}
\\[5mm]&=\left\{\begin{array}{lcl}
0 & \mbox{if} & s > k + 1
\\
{1 \over s} & \mbox{if} & s = k + 1
\\
\infty & \mbox{if} & s < k
\end{array}\right.
\end{align}
We see that the highest power of a polynomial must be $\ds{k + 1}$. For instance,
$\ds{\sum_{n\ =\ 0}n^{\dsc{2}}}$ is a polynomial of order $\dsc{2 + 1}=\dsc{3}$.

In addittion, the Stolz-Cesaro Theorem is quite helpful in dealing with the polynomial coefficients. We'll evaluate $\ds{A, B, C, D}$ with the present OP example:
$\ds{\,{\cal S}\pars{n}=\sum_{i\ =\ 0}^{n}i^{2}=An^{3} + Bn^{2} + Cn + D}$:
$$
\,{\cal S}\pars{0}=0\quad\imp\quad D = 0\quad\mbox{such that}\quad
\,{\cal S}\pars{n}=\sum_{i\ =\ 0}^{n}i^{2}=An^{3} + Bn^{2} + Cn
$$


$$
\mbox{Then,}\quad
\color{#66f}{\large A}
=\lim_{n\ \to \infty}{\,{\cal S}\pars{n} \over n^{3}}
=\color{#66f}{\large{ 1\over 3}}\quad\imp\quad
\,{\cal S}\pars{n} - {1 \over 3}\,n^{3}=Bn^{2} + Cn
$$

Again,
\begin{align}
\color{#66f}{\large B}&
=\lim_{n\ \to \infty}{\,{\cal S}\pars{n} - n^{3}/3 \over n^{2}} = 
\lim_{n\ \to\ \infty}\ {\pars{n + 1}^{2} - \pars{n + 1}^{3}/3 + n^{3}/3
\over 2n + 1} = \color{#66f}{\large\half}
\\[5mm]\imp&
\,{\cal S}\pars{n} - {1 \over 3}\,n^{3} - \half\,n^{2}=Cn
\end{align}

Finally,
  \begin{align}
\color{#66f}{\large C}&
=\lim_{n\ \to \infty}{\,{\cal S}\pars{n} - n^{3}/3 - n^{2}/2 \over n}
\\[5mm]&= \lim_{n\ \to\ \infty}\
\bracks{6\pars{n + 1}^{2} - 2\pars{n + 1}^{3} + 2n^{3}
- 3\pars{n + 1}^{2} + 3n^{2} \over 6}
= \color{#66f}{\large{1 \over 6}}
\end{align}

With $\ds{A = {1 \over 3}\,,\ B = \half}$ and $\ds{C = {1 \over 6}}$ we'll get:
$$
\color{#66f}{\large\,{\cal S}\pars{n}}
=\sum_{i\ =\ 0}^{n}i^{2}=An^{3} + Bn^{2} + Cn + D
=\color{#66f}{\large{1 \over 3}\,n^{3} + \half\,n^{2} + {1 \over 6}\,n}
$$
A: 
Why does $~\displaystyle\sum\limits_{i=1}^n i^2 = An^3+Bn^2+Cn + D$ ?

Why ? Imagine the Egyptians building the pyramids: On top of the pyramid, there is a single brick. Then, on the level directly below that, there are $2\times2=2^2$ bricks. Then, on the level directly below that, there are $3\times3=3^2$ bricks, etc. By the time we get at the base level, there are $n^2$ such bricks. So, how many bricks are there, in total ? Well, the volume of a smooth pyramid is the height times the basis over three, so $N\simeq\dfrac{n\cdot n^2}3=\dfrac{n^3}3$. Why only approximately? Because ours has edges, and is not smooth, their presence being responsible for the rest of the terms.
