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

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

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

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

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

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