just tell me if for r^2 it is {n(n1)(2n1)}/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
_________________________
n
Σ . ( r³ ) = n² (n+1)² / 4 = ( Σ r )²
r=1
__________________________
Σ . ( r² ) = n(n+1)(2n+1) / 6
r=1
_________________________
n
Σ . ( r³ ) = n² (n+1)² / 4 = ( Σ r )²
r=1
__________________________

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(n1)^3=1*[n^2+(n1)^2+n (n1)] =n^2+(n1)^2+n^2n =2*n^2+(n1)^2n 2^31^3=2*2^2+1^22 3^32^3=2*3^2+2^23 4^33^3=2*4^2+3^24 ...... n^3(n1)^3=2*n^2+(n1)^2n 各等式全相加 n^31^3=2*(2^2+3^2+... +n^2)+[1^2+2^2+...+(n1)^2] (2+3+4+...+n) n^31=2*(1^2+2^2+3^2+... +n^2)2+[1^2+2^2+... +(n1)^2+n^2]n^2(2+3+4+... +n) n^31=3*(1^2+2^2+3^2+... +n^2)2n^2(1+2+3+...+n)+1 n^31=3(1^2+2^2+...+n^2)1 n^2n(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)^4n^4=[(n+1)^2+n^2][(n +1)^2n^2] =(2n^2+2n+1)(2n+1) =4n^3+6n^2+4n+1 2^41^4=4*1^3+6*1^2+4*1+1 3^42^4=4*2^3+6*2^2+4*2+1 4^43^4=4*3^3+6*3^2+4*3+1 ...... (n+1)^4n^4=4*n^3+6*n^2+4*n +1 各式相加有 (n+1)^41=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)^41+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