$(a-b)^3=a^3-3a^2 b+3a b^2-b^3$ (remember the 3’s)
$(a+b)^3=a^3+3a^2 b+3a b^2+b^3$ (remember the 3’s)
$1^2 + 2^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}$
$1^3 + 2^3 + … + n^3 = \frac{n^2 (n+1)^2}{4}$
$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$
For an $m \times n$ grid of numbers satisfying:
- every row is an arithmetic progression left-to-right,
- every column is an arithmetic progression top-to-bottom,
let $a_{i,j}$ = the entry in row $i$, column $j$. Then there exist four constants $A, B, C, D$ such that $a_{i,j} = A + Bi + Cj + Dij$ holds for all $i, j$.