Công thức tổng dải số

$\sum _{k=0}^{n}{c}=nc$ where $c$ is some constant.

$\sum _{k=0}^{n}{k}={\frac {n(n+1)}{2}}$

$\sum _{k=0}^{n}{k^{2}}={\frac {n(n+1)(2n+1)}{6}}$

$\sum _{k=0}^{n}{k^{3}}={\frac {n^{2}(n+1)^{2}}{4}}$

$\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+\cdots =e^{x}$

$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots =\ln(1+x)\quad {\text{ for }}|x|<1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos(x)\quad {\text{ for all }}x\in \mathbb {C}$