Integration by summation

just tell me if for r^2 it is {n(n-1)(2n-1)}/6 then what would it be for r^3 because i need to solve for x^3.

n
Σ . ( r² ) = n(n+1)(2n+1) / 6
r=1
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n
Σ . ( r³ ) = n² (n+1)² / 4 = ( Σ r )²
r=1
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do you mean1^2+2^2+3^2+……+n^2=n(n+1) (2n+1)/6 ? if so, below is the answer: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-n 2^3-1^3=2*2^2+1^2-2 3^3-2^3=2*3^2+2^2-3 4^3-3^3=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)/2 3(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) 1^2+2^2+3^2+...+n^2=n(n+1) (2n+1)/6 1^3+2^3+3^3+……+n^3=[n(n +1)/2]^2 (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 2^4-1^4=4*1^3+6*1^2+4*1+1 3^4-2^4=4*2^3+6*2^2+4*2+1 4^4-3^4=4*3^3+6*3^2+4*3+1 ...... (n+1)^4-n^4=4*n^3+6*n^2+4*n +1 各式相加有 (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 4*(1^3+2^3+3^3+...+n^3)=(n +1)^4-1+6*[n(n+1)(2n+1)/6]+4* [(1+n)n/2]+n =[n(n+1)]^2 1^3+2^3+...+n^3=[n(n+1)/2]^2 just ignore the chinese character, it may help you to solve