# Solve the following

Question:

Given $a_{1}=\frac{1}{2}\left(a_{0}+\frac{A}{a_{0}}\right), a_{2}=\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{A}{a_{n}}\right)$ for $n \geq 2$, where $a>0, A$ $>0$.

Prove that $\frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right) 2^{n-1}$.

Solution:

Given: $a_{1}=\frac{1}{2}\left(a_{0}+\frac{A}{a_{0}}\right), a_{2}=\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{A}{a_{n}}\right)$ for $n \geq 2$, where $a>0, A$ $>0$.

To prove : $\frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{n-1}}$

Proof :

Let $\mathrm{p}(n): \frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{n-1}}$

Step I : For $n=1$,

$\mathrm{LHS}=\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}$

$R H S=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{1-1}}=\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}$

As, LHS = RHS

So, it is true for $n=1$.

Step II : For $n=k$,

Let $\mathrm{p}(k): \frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1}}$ be true for some values of $k \geq 2$.

Step III : For $n=k+1$,

$\mathrm{p}(k+1):$

$\mathrm{LHS}=\frac{a_{k+1}-\sqrt{A}}{a_{k+1}+\sqrt{A}}$

$=\frac{\frac{1}{2}\left(a_{k}+\frac{A}{a_{k}}\right)-\sqrt{A}}{\frac{1}{2}\left(a_{k}+\frac{A}{a_{k}}\right)+\sqrt{A}}$

$=\frac{\frac{1}{2}\left(\frac{a_{k}^{2}+A-2 a_{k^{\sqrt{A}}}}{a_{k}}\right)}{\frac{1}{2}\left(\frac{a_{k}^{2}+A+2 a_{k^{\sqrt{A}}}}{a_{k}}\right)}$

$=\frac{a_{k}{ }^{2}+A-2 a_{k} \sqrt{A}}{a_{k}{ }^{2}+A+2 a_{k} \sqrt{A}}$

$=\frac{\left(a_{k}-\sqrt{A}\right)^{2}}{\left(a_{k}+\sqrt{A}\right)^{2}}$

$=\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2}$

$=\left[\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1}}\right]^{2}$ (Using  step |II)

$=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1} \times 2}$

$=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1+1}}$

$=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k}}$

$\mathrm{RHS}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k+1-1}}$

$=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k}}$

As, LHS = RHS

So, it is also true for $n=k+1$.

Hence, $\frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{n-1}} .$