Question

lim x趋近与1,(x+x平方+...+x的n次方-n）/x-1=?

Answer 1

0/0型罗比塔法则,上下同求导

=lim(x -> 1)(1 + 2 +...+n) = n(n + 1)/2

或者化成

lim(x->1)((x - 1) + (x^2 - 1) + ...+ (x^n - 1))/(x - 1)

=lim(x - 1)(1 + (x + 1) + (x^2 + x + 1) +...+(x^(n - 1) + x^(n - 2) +...+1)) = 1 + 2 +..+ n = n(n + 1)/2