Q13 (b)

Criteria	Marks
Provides correct proof	4
Proves the inductive step, or equivalent merit	3
Establishes the base case and makes some progress with the inductive step, or equivalent merit	2
Establishes base case, or equivalent merit	1

Sample answer:
$a_1=\sqrt2,\quad a_2=\sqrt{2+\sqrt2},\quad a_3=\sqrt{2+\sqrt{2+\sqrt2}}\quad\ldots$
$a_{n+1}^2=2+a_n,\quad n\ge1$

Consider $n=1$ case:
$\text{RHS}=2\cos\frac\pi4=2\times\frac1{\sqrt2}=\sqrt2=a_1=\text{LHS}$
so the result is true for $n=1$.

Assume the result is true for $n=k$. Thus,
$a_k=2\cos\frac{\pi}{2^{k+1}},\quad k\ge1$

Prove the result is true for $n=k+1$:

$$\begin{aligned} a_{k+1}^2&=2+a_k\\ &=2+2\cos\frac\pi{2^{k+1}}\\ &=2\left[1+\cos\frac\pi{2^{k+1}}\right]\\ &=2\times2\cos^2\frac\pi{2^{k+2}}\\ &=4\cos^2\frac\pi{2^{k+2}} \end{aligned}$$

$\therefore a_{k+1}=2\cos\frac\pi{2^{k+2}}\quad(a_{k+1}>0)$
Thus if the result is true for $k$, it is also true for $k+1$. Hence, using the principle of mathematical induction, the result is true for all $k\ge1$.