WIKI

$$\begin{align} (1+x+x^2+x^3+\dots)&=\frac 1 {1-x}\\ (1+x^2+x^4+x^6+\dots)&=\frac 1 {1-x^2}\\ \end{align}$$

$$(1+x+x^2+x^3+x^4)=\frac{1-x^5}{1-x}$$

$$\begin{align} 数论a 为等比数列,首项 a_1 公比q\\ \sum\limits_{i=1}^na_i=\frac {a_1(1-q^n) }{1-q} \end {align}$$

$$\begin{align} (a+b)^n&=\sum\limits^n_{k=0}\binom{n}{k}a^kb^{n-k}\\ \binom{n}{k}&=\frac {P^k_n} {k!}=\frac{k!}{k!(n-k)!}\\ (1+x)^n&=\sum\limits^\infty_{k=0}\binom{n}{k}x^k\\ \frac 1 {(1-x)^n}&=\sum\limits^\infty_{k=0}\binom{n+k-1}{k}x^k\\\ e^x&=\sum\limits_{i=0}^\infty\frac 1{i!}x^i \end{align}$$