As Charlie rightly points out in the previous post, the formulas break down when x is 1 since they require division by x-1.  When x is 1, the answer is clearly n(n+1)/2.

Suppose now that x is not 1. Since

1+x+x^2+...+x^n=[x^(n+1)-1]/(x-1), we have

x^k+x^(k+1)+...+x^n=[x^(n+1)-x^k]/(x-1)=y_k, say. Thus the quantity we seek the sum of is 

 y_1+y_2+...+y_n={nx^(n+1)-y_1}/(x-1)

={nx^(n+1)-[x^(n+1)-x]/(x-1)}/(x-1)

=x[nx^(n+1)-(n+1)x^n+1]/(x-1)^2, the same result as previously obtained.