1^3+2^3^+3^3+.+n^3
1个回答
证明1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=[n(n+1)/2]^2 n^4-(n-1)^4 =[n^2-(n-1)^2][n^2+(n-1)^2] =(2n-1)(2n^2-2n+1) =4n^3-6n^2+4n-1 2^4-1^4=4*2^3-6*2^2+4*2-1 3^4-2^4=4*3^3-6*3^2+4*3-1 4^4-3^4=4*4^3-6*4^2+4*4-1 .n^4-(n-1)^4=4n^3-6n^2+4n-1 各等式全部相加 n^4-1^4=4*(2^3+3^3+...+n^3)-6*(2^2+3^2+...+n^2)+4(2+3+4+...+n)-(n-1) n^4-1^4=4*(1^3+2^3+3^3+...+n^3)-6*(1^2+2^2+3^2+...+n^2)+4(1+2+3+4+...+n)-(n-1)-2 n^4-1=4*(1^3+2^3+3^3+...+n^3)-6*n(n+1)(2n+1)/6+4*n(n+1)/2-n-1 n^4-1=4*(1^3+2^3+3^3+...+n^3)-n(n+1)(2n+1)+2n(n+1)-n-1 n^4-1=4*(1^3+2^3+3^3+...+n^3)-n(n+1)(2n+1)+2n(n+1)-n-1 4*(1^3+2^3+3^3+...+n^3) =n^4-1+n(n+1)(2n+1)-2n(n+1)+n+1 =n^4-1+(n+1)(2n^2-n)+n+1 =n^4-1+(2n^3+n^2-n)+n+1 =n^4+2n^3+n^2 =(n^2+n)^2 =(n(n+1))^2 1^3+2^3+3^3+...+n^3 =[n(n+1)/2]^2
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