Question

[Ko cần bít]

Tính 
\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}+2004\sqrt{2005}}\)

Answer 1

[Không Tên]

\(\forall n\inℕ^∗\)ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)

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

Thay n=1; n=2; n=3; .....; n=2004 Ta có:

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)

\(=1-\frac{1}{\sqrt{2005}}\)