The value of $\frac{18}{(5+\sqrt{7})} + \frac{4}{(\sqrt{7}+\sqrt{3})} + \frac{1}{(\sqrt{3}+\sqrt{2})} + \frac{1}{(\sqrt{2}+1)}$

Explanation

$$ \frac{18}{(5+\sqrt{7})} + \frac{4}{(\sqrt{7}+\sqrt{3})} + \frac{1}{(\sqrt{3}+\sqrt{2})} + \frac{1}{(\sqrt{2}+1)} $$ We know that– $$ = \frac{18(5-\sqrt{7})}{(5+\sqrt{7})(5-\sqrt{7})} + \frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})} + \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} + \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} $$ $$ = \frac{18(5-\sqrt{7})}{[5^2-(\sqrt{7})^2]} + \frac{4(\sqrt{7}-\sqrt{3})}{[(\sqrt{7})^2-(\sqrt{3})^2]} + \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} + \frac{\sqrt{2}-1}{[(\sqrt{2})^2-1^2]} $$ $$ [\because (a-b)(a+b) = a^2-b^2] $$ $$ = \frac{18(5-\sqrt{7})}{25-7} + \frac{4(\sqrt{7}-\sqrt{3})}{7-3} + \frac{\sqrt{3}-\sqrt{2}}{3-2} + \frac{\sqrt{2}-1}{2-1} $$ $$ = \frac{18(5-\sqrt{7})}{18} + \frac{4(\sqrt{7}-\sqrt{3})}{4} + \frac{(\sqrt{3}-\sqrt{2})}{1} + \frac{\sqrt{2}-1}{1} $$ $$ = 5 - \sqrt{7} + \sqrt{7} - \sqrt{3} + \sqrt{3} - \sqrt{2} + \sqrt{2} - 1 $$ $$ = 5 - 1 $$ $$ = 4 $$ (446) ## Page 45 Let, $$2 - \frac{1}{2 - \frac{1}{2 - \dots}}}$$ $$x = 2 - \frac{1}{2 - \frac{1}{2 - \dots}}}$$ Now, $$x = 2 - \frac{1}{x}$$ $$x = \frac{2x - 1}{x}$$ $$x^2 - 2x + 1 = 0$$ $$x^2 - x - x + 1 = 0$$ $$x(x - 1) - 1(x - 1) = 0$$ $$(x - 1)^2 = 0$$ $$x = 1$$ So, the value of the continued fraction is 1.