Question

prove that 1/1+√2 +1/√2+√3 +1/√3+√4 +1/√5+√6 +1/√6+√7 +1/√7+√8 +1/√8+√9
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Answer 1

$\mathtt\green{answer \: is \: in \: given \: attachment}$

Answer 2

$\tiny{{ \sf{ \dfrac{1}{1 + \sqrt{2} } + \dfrac{1}{ \sqrt{2 } + \sqrt{3} } + \dfrac{1}{\sqrt{3} + \sqrt{4} } + \dfrac{1}{ \sqrt{4} + \sqrt{5} } + \dfrac{1}{ \sqrt{5} + \sqrt{6} } + \dfrac{1}{ \sqrt{6} + \sqrt{7} } + \dfrac{1}{ \sqrt{7} + \sqrt{8} } + \dfrac{1}{ \sqrt{8} + \sqrt{9} } = 2}}} \\ \\ \\ \pmb{\tt{ \: Now \: Rationalize \: the \: L.H.S \: term}} \\ \tiny{{ \sf{ \dfrac{1( \sqrt{2} - 1) }{(1 + \sqrt{2} )( \sqrt{2} - 1)} + \dfrac{1( \sqrt{3} - \sqrt{2} )}{ (\sqrt{2 } + \sqrt{3} )( \sqrt{3} - \sqrt{2} )} + \dfrac{1( \sqrt{4} - \sqrt{3} )}{(\sqrt{3} + \sqrt{4})( \sqrt{4} - \sqrt{3}) } + \dfrac{1( \sqrt{5} - \sqrt{4}) }{ (\sqrt{4} + \sqrt{5} )( \sqrt{5} - \sqrt{4} ) } + \dfrac{1( \sqrt{6} - \sqrt{5} )}{ (\sqrt{5} + \sqrt{6} )( \sqrt{6} - \sqrt{5}) } + \dfrac{1( \sqrt{7} - \sqrt{6} )}{ (\sqrt{6} + \sqrt{7} )( \sqrt{7} - \sqrt{6}) } + \dfrac{1( \sqrt{8} - \sqrt{7} )}{( \sqrt{7} + \sqrt{8} )( \sqrt{8} - \sqrt{7} )} + \dfrac{1( \sqrt{9} - \sqrt{8} )}{ (\sqrt{8} + \sqrt{9} )( \sqrt{9} - \sqrt{8}) } }}} \\ \\ \small{ \pmb { \tt{Using \: formula = ( a - b ) \: ( a + b ) = a² - b²}}} \\ \small{{ \sf{ { \cancel{ \sqrt{ {2}}} } - 1 + { \cancel{ \sqrt{3} }}- { \cancel{\sqrt{2} }}+ { \cancel{\sqrt{4} }}- { \cancel{\sqrt{3} }}+ { \cancel{\sqrt{5}}} - { \cancel{ \sqrt{4} }}+ { \cancel{\sqrt{6} }} - { \cancel{\sqrt{5}}} +{ \cancel{ \sqrt{7} }} - { \cancel{ \sqrt{6}}} + { \cancel{\sqrt{8}}} - { \cancel{ \sqrt{7}}} + \sqrt{9} - { \cancel{\sqrt{8} }}}} }\\ { \sf{ \sqrt{9} - 1}} = 3 - 1 \\ { \sf{ \therefore \: L.H.S\: \: = \: {2}}} \\ \\ \bf L.H.S = R.H.S$