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 $\frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right) 2^{n-1}$
Let:
$\mathrm{P}(\mathrm{n}): \frac{\mathrm{a}_{\mathrm{n}}-\sqrt{\mathrm{A}}}{\mathrm{a}_{\mathrm{n}}+\sqrt{\mathrm{A}}}=\left(\frac{\mathrm{a}_{1}-\sqrt{\mathrm{A}}}{\mathrm{a}_{1}+\sqrt{\mathrm{A}}}\right)^{2^{\mathrm{n}-1}}$
Step1;
$\mathrm{P}(1):\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{1-1}} \quad$ which is true
$\mathrm{P}(2):\left(\frac{a_{2}-\sqrt{A}}{a_{2}+\sqrt{A}}\right)=\left(\frac{\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)-\sqrt{A}}{\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)+\sqrt{A}}\right)=\left(\frac{a_{1}+\frac{A}{a_{1}}-2 \sqrt{A}}{a_{1}+\frac{A}{a_{1}}+2 \sqrt{A}}\right)=\left(\frac{a_{1}+A-2 \sqrt{A}}{a_{1}+A+2 \sqrt{A}}\right)=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{2-1}}$
Thus, $\mathrm{P}(1)$ and $\mathrm{P}(2)$ are true.
Step 2:
Let $\mathrm{P} \quad \mathrm{k} \quad$ be true.
Now,
$\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1}} \quad \ldots \ldots(\mathrm{i})$
And,
$\mathrm{P}(\mathrm{k}+1): \frac{a_{k+1}-\sqrt{A}}{a_{k+1}+\sqrt{A}}=\frac{\frac{1}{2}\left(a_{k}+\frac{A}{a_{k}}\right)-\sqrt{A}}{\frac{1}{2}\left(a_{k}+\frac{A}{a_{k}}\right)+\sqrt{A}}$
$=\frac{\left(a_{k}+\frac{A}{a_{k}}\right)-2 \sqrt{A}}{\left(a_{k}+\frac{A}{a_{k}}\right)+2 \sqrt{A}}$
$=\frac{\left(\sqrt{a_{k}}-\sqrt{\frac{A}{a_{k}}}\right)^{2}}{\left(\sqrt{a_{k}}+\sqrt{\frac{A}{a_{k}}}\right)^{2}}$
$=\left(\frac{\sqrt{a_{k}}-\sqrt{\frac{A}{a_{k}}}}{\sqrt{a_{k}}+\sqrt{\frac{A}{a_{k}}}}\right)^{2}$
$=\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2}$
$=\left[\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1}}\right]^{2}$
$=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k}}$
$=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{(k+1)-1}}$
Thus, $P|k+1|$ is also true.
$=\frac{a_{k}\left(a_{k}+\sqrt{A}\right)\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2^{k-1}}-\sqrt{A}\left(a_{k}-\sqrt{A}\right)}{\left(a_{k}+\sqrt{A}\right)^{2}}$
$=\frac{a_{k}\left(a_{k}+\sqrt{A}\right)\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2^{k-1}}-\sqrt{A}\left(a_{k}+\sqrt{A}\right)\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2^{k-1}}}{\left(a_{k}+\sqrt{A}\right)^{2}}$[Using (i)]
$=\frac{\left(a_{k}+\sqrt{A}\right)\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2^{k-1}}\left(a_{k}-\sqrt{A}\right)}{\left(a_{k}+\sqrt{A}\right)^{2}}$
$=\frac{\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2^{k-1}}\left(a_{k}-\sqrt{A}\right)}{\left(a_{k}+\sqrt{A}\right)}$
$=\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{2^{k-1}}\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)$
$=\left(\frac{a_{k}-\sqrt{A}}{a_{k}+\sqrt{A}}\right)^{\left(2^{k-1}+1\right)}$
$=\left(\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right)^{2^{k-1}}\right)^{\left(2^{k-1}+1\right)}$