Question

谁会算（n+1)^2+(n+2)^2+(n+3)^2+.+(n+n)^2 的多少.超急!

Answer 1

=n^2+2n+1^2 +n^2+4n+2^2+n^2+6n+3^2+…+n^2+2n^2+n^2 =n^2 ×n +2n×（1+2+…+n）+[1^2+2^2+…+ n^2] =n^3+[2n×n(1+n)/2]+ [n(n+1)(2n+1)]/6 =n^3+n^2(1+n)+[(n^2+n)(2n+1)]/6 = n^3+ n^3+ n^2+[2 n^3+ n^2+ 2n^2+ n]/6 = 2n^3+ n^2+1/3×n^3+1/2×n^2+n/6 =7/3×n^3+ 3/2 ×n^2+n/6 附：1*2-1+2*3-2+3*4-3……+n(n+1)-n =[1*2+2*3+3*4+……+n(n+1)]-(1+2+3+……+n) =1/3(1*2*3-0*1*2)+1/3(2*3*4-1*2*3)+1/3(3*4*5-2*3*4)+……1/3[n*(n+1)(n+2)-(n-1)n(n+1)]-(1+2+3+……+n) =1/3[n(n+1)(n+2)]-[(n+1)n]/2 ==[n(n+1)(2n+1)]/6