﻿ Find the value of \dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+\dfrac{1}{100\sqrt{99}+99\sqrt{100}} 1/(2+sqrt(2))+1/(3sqrt(2)+2sqrt(3))+1/(4sqrt(3)+3sqrt(4))+...+1/(100sqrt(99)+99sqrt(100))

#### Question

Find the value of

\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}

1/(2+sqrt(2))+1/(3sqrt(2)+2sqrt(3))+1/(4sqrt(3)+3sqrt(4))+...+1/(100sqrt(99)+99sqrt(100))

Collected in the board: Calculate algebraic expressions

Steven Zheng posted 4 months ago

\dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1} }

=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1} }{[ (n+1)\sqrt{n}+n\sqrt{n+1} ][{(n+1)\sqrt{n}-n\sqrt{n+1} } ] }

=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)^2-n^2(n+1)}

=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n^3+2n^2+n-n^3-n^2}

=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}

=\dfrac{\sqrt{n} }{n} - \dfrac{\sqrt{n+1} }{n+1}

Therefore

\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}

=1-\dfrac{\sqrt{2} }{2}+\dfrac{\sqrt{2} }{2}-\dfrac{\sqrt{3} }{3}+ \dfrac{\sqrt{3} }{3} -\dfrac{\sqrt{4} }{4}+\dots+\dfrac{\sqrt{99} }{99}-\dfrac{\sqrt{100} }{100}

=1- \dfrac{\sqrt{100} }{100}

=\dfrac{9}{10}

Steven Zheng posted 4 months ago

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