Finite sum $\sum_{j=0}^{n-1} j^2$ How can I calculate this finite sum? Can someone help me?  

$$\sum_{j=0}^{n-1} j^2$$

 A: I like this explanation:

Each row in the first triangle sums to $j^2$
So the sum of the rows is the number we seek.
Take that triangle and rotate left and right.  The sum is of the three trianges, then is 3 times the number we seek.  But when we sum it up, we get $(2n+1)$ in every entry in the triange or $(2n+1)$ times the number of entries.
$3\sum_\limits{j = 1}^n j^2 = (2n+1)\sum_\limits{j = 1}^n j$
A: Notice
$$\begin{align}
j^2 
&= \frac12[ j(j+1) + (j-1)j ]\\
&= \frac12\left[\frac{\color{red}{(j+2)}-\color{green}{(j-1)}}{3}j(j+1)
+ \frac{\color{blue}{(j+1)}-\color{magenta}{(j-2)}}{3}(j-1)j\right]\\
&= \frac16\left[\left(
\underbrace{j(j+1)\color{red}{(j+2)}}_A - \underbrace{\color{green}{(j-1)}j(j+1)}_B
\right) + \left(
\underbrace{(j-1)j\color{blue}{(j+1)}}_C - \underbrace{\color{magenta}{(j-2)}(j-1)j}_D
\right)\right]\\
&= \frac16\left[
\underbrace{j(j+1)(2j+1)}_{A+C} - 
\underbrace{(j-1)j(2j-1)}_{B+D}\right]
\end{align}
$$
The sum at hand is a telescoping sum with
$$\sum_{j=1}^{n-1}j^2 = \frac{(n-1)n(2n-1)}{6}$$
A: You can prove by induction that $\sum\limits_{j=0}^{n-1}j^2=(2n^3-3n^2+n)/6$.

First, show that this is true for $n=1$:
$\sum\limits_{j=0}^{1-1}j^2=(2-3+2)/6$
Second, assume that this is true for $n$:
$\sum\limits_{j=0}^{n-1}j^2=(2n^3-3n^2+n)/6$
Third, prove that this is true for $n+1$:
$\sum\limits_{j=0}^{n}j^2=$
$\color\red{\sum\limits_{j=0}^{n-1}j^2}+n^2=$
$\color\red{(2n^3-3n^2+n)/6}+n^2=$
$[2(n+1)^3-3(n+1)^2+(n+1)]/6$

Please note that the assumption is used only in the part marked red.
A: Try to break a cube into L-shells, then break each L-shell into $1$, $n$, and $n^2$ components.

A: The sum of $k$th powers is described by a polynomial of degree $k+1$, and can be found by interpolation. Just for fun, let's cheat a bit to make the interpolation easy.
It would be nice not to have to calculate powers of large numbers, so let's use $n=2$ as an upper limit, for which we have $\sum_{j=0}^{n-1}j^2 = 1$. Extending the sum backwards:
$$\begin{array}{c|c}
n & ``\sum_{j=0}^{n-1}j^2"\\
\hline
2 & 1\\
1 & 0\\
0 & 0\\
-1 & -1\\
\end{array}
$$
(The cheating is that of course the sum doesn't work that way for $n=-1$. Instead, the entries in the second column are calculated by successively subtracting squares from the entry above.)
Now we have very simple values for solving the coefficients in $An^3 + Bn^2 + Cn + D$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\left.\begin{array}{rclcl}
\ds{j^{\underline{2}}} & \ds{=} & \ds{j\pars{j - 1}} & \ds{=} & \ds{j^{2} - j}
\\[1mm]
\ds{j^{\underline{1}}} & \ds{=} & \ds{j}
\end{array}\right\}
\implies
j^{2} = j^{\underline{2}} + j^{\underline{1}}}$

\begin{align}
\sum_{j = 0}^{n - 1}j^{2} & =
\sum_{j = 0}^{n - 1}\pars{j^{\underline{2}} + j^{\underline{1}}} =
{1 \over 3}\,n^{\underline{3}} + {1 \over 2}\,n^{\underline{2}} =
{1 \over 3}\,n\pars{n - 1}\pars{n - 2} + {1 \over 2}\,n\pars{n - 1}
\\[5mm] & =
n\pars{n - 1}\pars{{n - 2 \over 3} + {1 \over 2}} =
\bbx{\ds{n\pars{n - 1}\pars{2n - 1} \over 6}}
\end{align}
A: This problem has many answers:
HINT : We define 
$a_{n-1}=$$\sum_{j=0}^{n-1} j^2$ So we need to solve the following recurrent sequence
$a_{n}=$ $a_{n-1}+n^2$
