结论：1+2+⋯+n=(1+n)n21+2+\cdots+n=(1+n)\frac n21+2+⋯+n=(1+n)2n
证明：经典的高斯求和
结论：12+22+⋯+n2=16n(n+1)(2n+1)1^2+2^2+\cdots+n^2=\frac16n(n+1)(2n+1)12+22+⋯+n2=61n(n+1)(2n+1)
证明：(n+1)3=n3+3n2+3n+1{(n+1)3−n3=3n2+3n+1n3−(n−1)3=3(n−1)2+3(n−1)+1⋯⋯⋯⋯33−23=3⋅22+3⋅2+123−13=3⋅12+3⋅1+1全部相加并整理后得解(n+1)^3=n^3+3n^2+3n+1\\ \left\{\begin{aligned} (n+1)^3-n^3&=3n^2+3n+1\\ n^3-(n-1)^3&=3(n-1)^2+3(n-1)+1\\ \cdots\cdots&\cdots\cdots\\ 3^3-2^3&=3\cdot2^2+3\cdot2+1\\ 2^3-1^3&=3\cdot1^2+3\cdot1+1\\ \end{aligned}\right.\\ 全部相加并整理后得解(n+1)3=n3+3n2+3n+1⎩⎨⎧(n+1)3−n3n3−(n−1)3⋯⋯33−2323−13=3n2+3n+1=3(n−1)2+3(n−1)+1⋯⋯=3⋅22+3⋅2+1=3⋅12+3⋅1+1全部相加并整理后得解
