Base case: For \(n = 1\),

\(1 = \frac{1 - 2x + x^2}{(1-x)^2} = \frac{(1-x)^2}{(1-x)^2},\)

so \(H_1\) is true.

Inductive step: Assume that

\(\sum_{r=1}^{k} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2}.\)

Consider \(\sum_{r=1}^{k+1} r x^{r-1}\):

\(\sum_{r=1}^{k+1} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2} + (k+1)x^k.\)

Combine over a common denominator:

\(\frac{1 - (k+1)x^k + kx^{k+1} + (k+1)x^k(1-2x+x^2)}{(1-x)^2}.\)

Simplify:

\(\frac{1 + kx^{k+1} + (k+1)x^k(-2x + x^2)}{(1-x)^2} = \frac{1 - (k+2)x^{k+1} + (k+1)x^{k+2}}{(1-x)^2}.\)

Thus, \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).