试证明1^3+2^3+3^3+…+n^3=（1+2+3+…+n）^2

证明,方法一：(n+1)^4-n^4=4n^3+6n^2+4n+1.∴n^3=(1/4)[(n+1)^4-n^4]-(3/2)n^2-n-1/4∴左边=∑i^3=(1/4)[(n+1)^4-1]-(3/2)*(1/6)n(n+1)(2n+1)-(1/4)n-(n+1)n/2=(1/4)(n^4+4n^3+6n^2+4n-2n^3-3n^2-n-n)-(1/2)(n^2+n)=(1/4)(n^4+2n^3+n^2)=[(1/2)n(n+1)]^2=(1+2+3+…+n)^2[附注:这里用了另一个公式∑i^2=(1/6)n(n+1)(2n+1)证明如下:(n+1)^3-n^3=3n^2+3n+1 ∴n^2=(1/3)[(n+1)^3-n^3]-n-1/3∴∑i^2=(1/3)[(n+1)^3-1]-(1/2)n(n+1)-n/3=.=(1/6)n(n+1)(2n+1)]方法二：数学归纳法当n=1时,左边1^3=1,右边1^2=1左边=右边假设当n=k时等式成立 1^3+2^3+3^3+…k^3=（1+2+3+.+k)^2则当n=k+1时1^3+2^3+3^3+…k^3+(k+1)^3=(1+2+3+.+k)^2+(k+1)^3 1+2+3.+k=k(k+1)/2 等差数列=k^2(1+k)^2/4+(k+1)^3 =(1+k)^2(k^2/4+k+1)=(1+k)^2(k^2+4k+4)/4=(k+1)^2(k+2)^2/4=[(k+1)(k+1+1)/2]^2=(1+2+3.+k+k+1)^2 1+2+3+...k+k+1=(k+1)(k+1+1)/2 也是等差数列所以当n=k+1等式也成立所以,1^3+2^3+3^3+.+n^3=（1+2+3+.+n)^2

1^3=1^2
1^3 2^3=9=（1 2）^2
1^3 2^3 3^3=36=（1 2 3）^2
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k^3
=1/4*4k*k^2
=1/4[(k+1)^2-(k+1)^2]*k^2
=[k(k+1)/2]^2-[k(k-1)/2]^2

1^3+2^3+3^3+…+n^3
=1-0+9-1+……+[n(n+1)/2]^2-[n(n-1)/2]^2
=[n(n+1)/2]^2
1+2+3+…+n=n(n+1)/2