1、1到N的平方和推导:1²+2²+3²+。。。+n²=n(n+1)(2n+1)/6
由1²+2²+3²+。。。+n²=n(n+1)(2n+1)/6
∵(a+1)³-a³=3a²+3a+1(即(a+1)³=a³+3a²+3a+1)
a=1时:2³-1³=3×1²+3×1+1
a=2时:3³-2³=3×2²+3×2+1
a=3时:4³-3³=3×3²+3×3+1
a=4时:5³-4³=3×4²+3×4+1
......
a=n时:(n+1)³-n³=3×n²+3×n+1
等式两边相加:
(n+1)³-1=3(1²+2²+3²+。。。+n²)+3(1+2+3+。。。+n)+(1+1+1+。。。+1)
3(1²+2²+3²+。。。+n²)=(n+1)³-1-3(1+2+3+。。。+n)-(1+1+1+。。。+1)
3(1²+2²+3²+。。。+n²)=(n+1)³-1-3(1+n)×n÷2-n
6(1²+2²+3²+。。。+n²)=2(n+1)³-3n(1+n)-2(n+1)
=(n+1)[2(n+1)²-3n-2]
=(n+1)[2(n+1)-1][(n+1)-1]
=n(n+1)(2n+1)
∴1²+2²+。。。+n²=n(n+1)(2n+1)/6
2、1到N的立方和推导:1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2
推导: (n+1)^4-n^4=4n^3+6n^2+4n+1,
n^4-(n-1)^4=4(n-1)^3+6(n-1)^2+4(n-1)+1,
......
2^4-1^4=4*1^3+6*1^2+4*1+1,
把这n个等式两端分别相加,得:
(n+1)^4-1=4(1^3+2^3+3^3...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+3+...+n)+n
由于1+2+3+...+n=(n+1)n/2,
1^2+2^2+...+n^2=n(n+1)(2n+1)/6,
代人上式整理后得:
1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2