求证：1*2+2*5+3*8+…+n(3n-1)=n^2(n+1)

题目

求证：1*2+2*5+3*8+…+n(3n-1)=n^2(n+1)

答案

an=n(3n-1)=3n^2-n
sn=[3*1^2-1]+[3*2^2-2]+[3*3^2-3]+……+[3(n-1)^2-(n-1)]+[3n^2-n]
=[3*1^2+3*2^2+3*3^2+……+3(n-1)^2+3n^2]-[1+2+3+……+(n-1)+n]
=3n(n+1)(2n+1)/6-(n+1)n/2
=n(2n^2+2n)/2
=n^2*(n+1)