CAIE A-Level Further Math AS 1.7 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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$\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{2} \mathrm{e}^{x}\right)=x^{2} \mathrm{e}^{x}+2 x \mathrm{e}^{x}=\left(x^{2}+2 x\right) \mathrm{e}^{x}$ so true when n=1.
Differentiates once using the product
rule.

Assume that $\frac{\mathrm{d}^{k}}{\mathrm{~d} x^{k}}\left(x^{2} \mathrm{e}^{x}\right)=\left(x^{2}+2 k x+k(k-1)\right) \mathrm{e}^{x}$
[for some value of k ].
States inductive hypothesis.

$\frac{\mathrm{d}^{k+1}}{\mathrm{~d} x^{k+1}}\left(x^{2} \mathrm{e}^{x}\right)=\left(x^{2}+2 k x+k(k-1)\right) \mathrm{e}^{x}+\mathrm{e}^{x}(2 x+2 k)$
Differentiates k th derivative.

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

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