由求和公式,S(k) = k*a(1) + k(k-1)d / 2又k(k-1) = [ k³ - (k-1)³ - 1] / 3所以S(1) + S(2) + ...+ S(n) = (1+2+...+n)a(1) + (1*2 + 2*3 + ...+(n-1)n) d / 2= n(n+1)a(1) / 2 + (n³ - 1 - n) / 3 ...