Užitečné sumarizační vzorce

$\displaystyle\sum_{i=l}^u = u - l + 1$

$\displaystyle\sum_{i=1}^n 1 = n$

$\displaystyle\sum_{i=1}^n i = 1 + 2 + ... + n = \dfrac{1}{2}n(n+1)$

$\displaystyle\sum_{i=1}^n i^2 = 1^2 + 2^2 + ... + n^2 = \dfrac{1}{6}n(n+1)(2n+1)\approx\dfrac{1}{3}n^3$

$\displaystyle\sum_{i=0}^n a^i= 1 + a + a^2 + ... + a^n = \dfrac{a^{n+1}-1}{a-1}$, pro $a \not= 1$

$\displaystyle\sum_{i=0}^n 2^i = 2^0 + 2^1 + ... 2^n = 2^{n+1}-1$