bmai68191 bmai68191
  • 19-10-2021
  • Mathematics
contestada

[tex]\frac{1}{1+\sqrt{2} } +\frac{1}{\sqrt{2} +\sqrt{3} } +...+\frac{1}{\sqrt{99} +\sqrt{100} }[/tex]

Respuesta :

evgeniylevi
evgeniylevi evgeniylevi
  • 19-10-2021

Answer:

9

Step-by-step explanation:

the given expression can be re-written:

[tex]\frac{\sqrt{2} -1}{(\sqrt{2}-1)(\sqrt{2}+1)} +\frac{\sqrt{3} -\sqrt{2}}{(\sqrt{3} +\sqrt{2})(\sqrt{3} -\sqrt{2})} + ...+\frac{\sqrt{100} -\sqrt{99}}{(\sqrt{100} -\sqrt{99})(\sqrt{100} +\sqrt{99})};[/tex]

then

[tex]\sqrt{2} -1+\sqrt{3} -\sqrt{2} +\sqrt{4} -\sqrt{3}+\sqrt{5} -\sqrt{4}+...+\sqrt{100}-\sqrt{99}= -1+\sqrt{100}=9.[/tex]

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