1.2.4 特殊的有限级数

$$\begin{array}{}\text{(1.55)}& 1+2+3+\cdots +\left(n-1\right)+n=\frac{n\left(n+1\right)}{2},\end{array}$$$$\begin{array}{}\text{(1.56)}& p+\left(p+1\right)+\left(p+2\right)+\cdots +\left(p+n\right)=\frac{\left(n+1\right)\left(2p+n\right)}{2},\end{array}$$$$\begin{array}{}\text{(1.57)}& 1+3+5+\cdots +\left(2n-3\right)+\left(2n-1\right)=n^2,\end{array}$$$$\begin{array}{}\text{(1.58)}& 2+4+6+\cdots +\left(2n-2\right)+2n=n\left(n+1\right),\end{array}$$$$\begin{array}{}\text{(1.59)}& 1^2+2^2+3^2+\cdots +{(n-1)}^2+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6},\end{array}$$$$\begin{array}{}\text{(1.60)}& 1^3+2^3+3^3+\cdots +{(n-1)}^3+n^3=\frac{n^2{(n+1)}^2}{4},\end{array}$$$$\begin{array}{}\text{(1.61)}& 1^2+3^2+5^2+\cdots +{(2n-1)}^2=\frac{n\left(4n^2-1\right)}{3},\end{array}$$$$\begin{array}{}\text{(1.62)}& 1^3+3^3+5^3+\cdots +{(2n-1)}^3=n^2\left(2n^2-1\right),\end{array}$$$$\begin{array}{}\text{(1.63)}& 1^4+2^4+3^4+\cdots +n^4=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30},\end{array}$$$$\begin{array}{}\text{(1.64)}& 1+2x+3x^2+\cdots +n{x}^{n-1}=\frac{1-\left(n+1\right)x^n+n{x}^{n+1}}{{(1-x)}^2}\left(x\ne 1\right).\end{array}$$