已知数列{an}的通项公式an=n²+n,则{an}的前n项和Sn=?

an =n^2+n
= n(n+1)
= (1/3)[n(n+1)(n+2) - (n-1)n(n+1)]
Sn = (1/3){ summation (i:1->n)[i(i+1)(i+2) - (i-1)i(i+1)] }
= (1/3)n(n+1)(n+2)