Qs. No. 4...??

Dear student,

Kindly find the answer to your query:
$$\begin{gathered} \frac{1}{\sqrt{3+2\sqrt{2}}}+\frac{1}{\sqrt{5+2\sqrt{6}}}+\frac{1}{\sqrt{7+2\sqrt{12}}}+.....+\frac{1}{\sqrt{199+2\sqrt{9900}}} \\ =\frac{1}{\sqrt{1^2+{\left(\sqrt{2}\right)}^2+2\times 1\times \sqrt{2}}}+\frac{1}{\sqrt{{\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2+2\times \sqrt{3}\times \sqrt{2}}}+\frac{1}{\sqrt{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{4}\right)}^2+2\times \sqrt{3}\times \sqrt{2}}}+.....+\frac{1}{\sqrt{{\left(\sqrt{100}\right)}^2+{\left(\sqrt{99}\right)}^2+2\times \sqrt{100}\times \sqrt{99}}} \\ =\frac{1}{\sqrt{{\left(\sqrt{2}+1\right)}^2}}+\frac{1}{\sqrt{{\left(\sqrt{3}+\sqrt{2}\right)}^2}}+\frac{1}{\sqrt{{\left(\sqrt{4}+\sqrt{3}\right)}^2}}+.....+\frac{1}{\sqrt{{\left(\sqrt{100}+\sqrt{99}\right)}^2}} \\ =\frac{1}{\left(\sqrt{2}+1\right)}+\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\left(\sqrt{4}+\sqrt{3}\right)}+.....+\frac{1}{\left(\sqrt{100}+\sqrt{99}\right)} \\ =\frac{1}{\left(\sqrt{2}+1\right)}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{1}{\left(\sqrt{4}+\sqrt{3}\right)}\times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}}+.....+\frac{1}{\left(\sqrt{100}+\sqrt{99}\right)}\times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}} \\ =\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+.........+\frac{\sqrt{100}-\sqrt{99}}{100-99} \\ =\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+....+\sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ \mathrm{We} \mathrm{can} \mathrm{see} \mathrm{that} \mathrm{terms} \mathrm{will} \mathrm{get} \mathrm{cancelled} \mathrm{and} \mathrm{we} \mathrm{are} \mathrm{left} \mathrm{with} \\ =-1+\sqrt{100} \\ =9 \end{gathered}$$

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