Question

Evaluate:

$\left(i\right)\frac{16.2^{n+1}-4.2^n}{16.2^{n+2}-2.2^{n+2}}$

$\left(ii\right)\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}$

Answer 1

(i) $\frac{{16.2}^{n+1}-{4.2}^n}{{16.2}^{n+2}-{2.2}^{n+2}}$

Let $a=\frac{{16.2}^{n+1}-{4.2}^n}{{16.2}^{n+2}-{2.2}^{n+2}}$

$=\frac{2^{4+n+1}-2^{2+n}}{2^4{.2}^{n+2}-2^{1+n+2}}\dots (2^4=16)$

$=\frac{2^{n+5}-2^{n+2}}{2^{n+6}-2^{n+3}}$

$=\frac{2^{n+2}(2^3-1)}{2^{n+3}(2^3-1)}$

$=2^{n+2-n-3}$

$a=2^{-1}=\frac{1}{2}$

(ii) $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}$

Let $b=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}$

$=\frac{1-\sqrt{2}}{1-(\sqrt{2}{)}^2}+\frac{\sqrt{-2}-\sqrt{3}}{(\sqrt{2}{)}^2-(\sqrt{3}{)}^2}+\frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}{)}^2-(\sqrt{4}{)}^2}\dots \left(Rationalizing\right)$

$=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}$

$=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}$

$b=-1+2=1$.