Question

1平方+2平方+3平方+4平方+.+n平方的和 求证明

Answer 1

答案是：n（n+1）（2n+1）/6 解析：(n+1)^3-n^3=3n^2+3n+1 n^3-(n-1)^3=3(n-1)^2+3(n-1)+1 (n+1)^3-1=3*(1^2+2^2+……+n^2)+3*(1+2+……+n)+n*1 (n+1)^3-1=3*(1^2+2^2+……+n^2)+3*n(n+1)/2+n 1^2+2^2+……+n^2=[(n+1)^3-3n(n+1)/2-(n+1)]/3 =(n+1)(n^2+2n+1-3n/2-1)/3 =(n+1)(2n^2+n)/6 =n(n+1)(2n+1)/6