精品文档---下载后可任意编辑自然数平方数列和立方数列求和公式怎么推导即：(1) 1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6(2) 1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2推导过程如下：一
1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6 利用立方差公式n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]=n^2+(n-1)^2+n^2-n=2*n^2+(n-1)^2-n2^3-1^3=2*2^2+1^2-23^3-2^3=2*3^2+2^2-34^3-3^3=2*4^2+3^2-4
n^3-(n-1)^3=2*n^2+(n-1)^2-n各等式全相加n^3-1^3=2*(2^2+3^2+
+n^2)+[1^2+2^2+
+(n-1)^2]-(2+3+4+
+n)n^3-1=2*(1^2+2^2+3^2+
+n^2)-2+[1^2+2^2+
+(n-1)^2+n^2]-n^2-(2+3+4+
+n) n^3-1=3*(1^2+2^2+3^2+
+n^2)-2-n^2-(1+2+3+
+n)+1 n^3-1=3(1^2+2^2+
+n^2)-1-n^2-n(n+1)/23(1^2+2^2+
+n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)=(n/2)(n+1)(2n+1)故：1^2+2^2+3^2+
+n^2=n(n+1)(2n+1)/6二
1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2 证明如下：(n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]=(2n^2+2n+1)(2n+1)=4n^3+6n^2+4n+1精品文档---下载后可任意编辑2^4-1^4=4*1^3
