Question

Va rog frumos urgent dau coroana ​

Answer 1

Răspuns:

Explicație pas cu pas:

Amplificăm fiecare fracție cu conjugata numitorului (pentru a raționaliza numitorii):

$S=\dfrac{1}{\sqrt{2} +1} +\dfrac{1}{\sqrt{3} +\sqrt{2} }+\dfrac{1}{\sqrt{4} +\sqrt{3} }+...+\dfrac{1}{\sqrt{n+1} +\sqrt{n} }=\dfrac{1*(\sqrt{2} -1)}{(\sqrt{2} -1)(\sqrt{2}+1) } +\\+\dfrac{1*(\sqrt{3} -\sqrt{2} )}{(\sqrt{3} -\sqrt{2} )(\sqrt{3}+\sqrt{2} ) }+\dfrac{1*(\sqrt{4} -\sqrt{3} )}{(\sqrt{4} -\sqrt{3} )(\sqrt{4}+\sqrt{3} ) }+...+\dfrac{1*(\sqrt{n+1} -\sqrt{n} )}{(\sqrt{n+1} -\sqrt{n} )(\sqrt{n+1}+\sqrt{n} ) }=$$=\dfrac{\sqrt{2} -1}{(\sqrt{2})^2 -1^2}+\dfrac{\sqrt{3} -\sqrt{2} }{(\sqrt{3})^2 -(\sqrt{2}) ^2}+\dfrac{\sqrt{4} -\sqrt{3} }{(\sqrt{4})^2 -(\sqrt{3}) ^2}+...+\dfrac{\sqrt{n+1} -\sqrt{n} }{(\sqrt{n+1})^2 -(\sqrt{n}) ^2}=\\=\dfrac{\sqrt{2} -1 }{2-1}+\dfrac{\sqrt{3} -\sqrt{2}}{3-2}+\dfrac{\sqrt{4} -\sqrt{3}}{4-3}+...+\dfrac{\sqrt{n+1} -\sqrt{n}}{n+1-n}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1.\\ Deci,~S=\sqrt{n+1}-1.$

$b)~~S=2018,~=>~\sqrt{n+1}-1=2018,~=>~\sqrt{n+1}=2018+1,~=>~\sqrt{n+1}=2019~|^2,~=>~n+1=2019^2,~=>~n=2019^2-1.\\Raspuns:~n=2019^2-1.$