Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Lê Tài Bảo Châu]

Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n.\sqrt{n+1}}=\frac{1}{\sqrt{n.\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n.\left(n+1\right)}.\left(n+1-n\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n.\left(n+1\right)}}\)

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

\(T=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=1-\frac{1}{\sqrt{100}}=\frac{9}{10}\)

b) \(\Leftrightarrow1-\frac{1}{\sqrt{n+1}}=\frac{4}{5}\)( làm tắt nhé )

\(\Leftrightarrow\frac{1}{\sqrt{n+1}}=\frac{1}{5}\)

\(\Leftrightarrow\sqrt{n+1}=5\)

\(\Leftrightarrow n=24\)

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