Question

simplify

1/1+root2 + 1/root2+root3 + 1/root3+root4 ..... +1/root8+root9


please explain how to do it I am tierd now

Answer 1

hope the attachment will help u

Answer 2

Given:

$\frac{1}{1+\sqrt 2}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}$

To find:

Value of $\frac{1}{1+\sqrt 2}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}$

Solution:

By rationalization

$\frac{1}{1+\sqrt 2}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}\times \frac{\sqrt{9}-\sqrt{8}}{\sqrt{9}-\sqrt{8}}$

$\frac{\sqrt{2}-1}{(\sqrt{2})^2-1}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4})^2-(\sqrt{3})^2}+....\frac{\sqrt{9}-\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}$

Using identity:

$(a+b)(a-b)=a^2-b^2$

$\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+\frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{\sqrt{7}-\sqrt{6}}{7-6}+\frac{\sqrt{8}-\sqrt{7}}{8-7}+\frac{\sqrt{9}-\sqrt{8}}{9-8}$

$\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}$

$=-1+3$

$=2$

$\frac{1}{1+\sqrt 2}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}=2$