CAIE A-Level Further Math AS 1.7.1 Proof By Induction Questions

$1=\frac{1-2 x+x^{2}}{(1-x)^{2}}=\frac{(1-x)^{2}}{(1-x)^{2}}$ so $\mathrm{H}_{1}$ is true.
Checks base case.

Assume that $\sum_{r=1}^{k} r x^{r-1}=\frac{1-(k+1) x^{k}+k x^{k+1}}{(1-x)^{2}}$.
States inductive hypothesis.

$\sum_{r=1}^{k+1} r x^{r-1}=\frac{1-(k+1) x^{k}+k x^{k+1}}{(1-x)^{2}}+(k+1) x^{k}$
Considers sum to k+1.

$\frac{1-(k+1) x^{k}+k x^{k+1}+(k+1) x^{k}\left(1-2 x+x^{2}\right)}{(1-x)^{2}}$
Puts over a common denominator.

$\frac{1+k x^{k+1}+(k+1) x^{k}\left(-2 x+x^{2}\right)}{(1-x)^{2}}=\frac{1-(k+2) x^{k+1}+(k+1) x^{k+2}}{(1-x)^{2}}$
So $\mathrm{H}_{k+1}$ is true. By induction, $\mathrm{H}_{n}$ is true for all positive integers n.
States conclusion.

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