Question

Prove that

\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

Collected in the board: Square Root

Steven Zheng posted 3 weeks ago


Answer

\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}}

=\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}

=\sqrt{9}-1

=3-1=2


Steven Zheng posted 3 weeks ago

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