Question

tìm n sao cho

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)

giúp em với

Answer 1

Có \(\frac{1}{\sqrt{1}+\sqrt{2}}\)+ \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + ... + \(\frac{1}{\sqrt{n-1}+\sqrt{n}}\)= 11

<=> -1+ \(\sqrt{2}\)- \(\sqrt{2}\)+ \(\sqrt{3}\)-...- \(\sqrt{n-1}\)+ \(\sqrt{n}\)= 11

<=> \(\sqrt{n-1}\)= 11

<=> \(\sqrt{n}\) = 11 + 1 = 12

<=> n = 144

Vậy n = 144 thì \(\frac{1}{\sqrt{1}+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + ... + \(\frac{1}{\sqrt{n-1}+\sqrt{n}}\) = 11