Prove that: 1 1+2 +1 2+3 +1 3+4 +1 4+5 +1 5+6 +1 6+7 +1 7+8 +1 8+9 =2

L.H.S.=1 1+2 +1 2+3 +1 3+4 +1 4+5 +1 5+6 +1 6+7 +1 7+8 +1 8+9 =1 1+2 1-2 1-2 +1 2+3 2-3 2-3 +1 3+4 3-4 3-4 +1 4+5 4-5 4-5 +1 5+6 5-6 5-6 +1 6+7 6-7 6-7 +1 7+8 7-8 7-8 +1 8+9 8-9 8-9 = 1-2 1- (2 )^2 + 2-3 (2 )^2- (3 )^2 + 3-4 (3 )^2- (4 )^2 + 4-5 (4 )^2- (5 )^2 + 5-6 (5 )^2- (6 )^2 + 6-7 (6 )^2- (7 )^2 + 7-8 (7 )^2- (8 )^2 + 8-9 (8 )^2- (9 )^2 = 1-2 1-2 + 2-3 2-3 + 3-4 3-4 + 4-5 4-5 5-6 5-6 + 6-7 6-7 + 7-8 7-8 + 8-9 8-9 = 1-2 -1 + 2-3 -1 + 3-4 -1 + 4-5 -1 5-6 -1 + 6-7 -1 + 7-8 -1 + 8-9 -1 =-1+2-2+3-3+4-4+5-5 +6-6+7-7+8-8+9-9 =-1+3 =2 =R.H.S.

Prove that: 1 1+2 +1 2+3 +1 3+4 +1 4+5 +1 5+6 +1 6+7 +1 7+8 +1 8+…

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Question

Prove that:
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\\+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}=2$

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Answer

$\text{L.H.S.}=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\\+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}$
$=\frac{1}{1+\sqrt{2}}\times\frac{1-\sqrt{2}}{1-\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\times\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}\\\times\frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\times\frac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\times\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}\\+\frac{1}{\sqrt{6}+\sqrt{7}}\times\frac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}\times\frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\times\frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}$
$=\frac{1-\sqrt{2}}{1-\big(\sqrt{2}\big)^2}+\frac{\sqrt{2}-\sqrt{3}}{\big(\sqrt{2}\big)^2-\big(\sqrt{3}\big)^2}+\frac{\sqrt{3}-\sqrt{4}}{\big(\sqrt{3}\big)^2-\big(\sqrt{4}\big)^2}\\+\frac{\sqrt{4}-\sqrt{5}}{\big(\sqrt{4}\big)^2-\big(\sqrt{5}\big)^2}+\frac{\sqrt{5}-\sqrt{6}}{\big(\sqrt{5}\big)^2-\big(\sqrt{6}\big)^2}+\frac{\sqrt{6}-\sqrt{7}}{\big(\sqrt{6}\big)^2-\big(\sqrt{7}\big)^2}\\+\frac{\sqrt{7}-\sqrt{8}}{\big(\sqrt{7}\big)^2-\big(\sqrt{8}\big)^2}+\frac{\sqrt{8}-\sqrt{9}}{\big(\sqrt{8}\big)^2-\big(\sqrt{9}\big)^2}$
$=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\\\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9}$
$=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\\\frac{\sqrt{5}-\sqrt{6}}{-1}+\frac{\sqrt{6}-\sqrt{7}}{-1}+\frac{\sqrt{7}-\sqrt{8}}{-1}+\frac{\sqrt{8}-\sqrt{9}}{-1}$
$=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-\sqrt{4}+\sqrt{5}-\sqrt{5}\\+\sqrt{6}-\sqrt{6}+\sqrt{7}-\sqrt{7}+\sqrt{8}-\sqrt{8}+\sqrt{9}-\sqrt{9}$
$=-1+3$
$=2$
$=\text{R.H.S.}$