lim(x→1) (x+x2+.+xn-n)/(x-1)= lim(x→1) [x^(n+1)-x-n(x-1)]/(x-1)^2 (0/0型,运用洛必达法则）= lim(x→1)1/2* [(n+1)x^n-1-n]/(x-1) (0/0型,运用洛必达法则）=lim(x→1)1/2* n(n+1)x^(n-1)=n(n+1)/2...