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

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


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.
