Đáp án:
`A=1-\sqrt{2005}/{2005}`
Giải thích các bước giải:
Với mọi `n\in NN`*; ta có:
`\qquad 1/{(n+1)\sqrt{n}+n\sqrt{n+1}}`
`=1/{\sqrt{n(n+1)}.(\sqrt{n+1}+\sqrt{n})}`
`={\sqrt{n+1}-\sqrt{n}}/{\sqrt{n(n+1)}.(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}`
`={\sqrt{n+1}-\sqrt{n}}/{\sqrt{n(n+1)}(n+1-n)}`
`={\sqrt{n+1}-\sqrt{n}}/\sqrt{n(n+1)}`
`=\sqrt{n+1}/\sqrt{n(n+1)}-\sqrt{n}/\sqrt{n(n+1)}``=1/\sqrt{n}-1/\sqrt{n+1}`
`=> 1/{(n+1)\sqrt{n}+n\sqrt{n+1}}=1/\sqrt{n}-1/\sqrt{n+1}` với `n\in N`* `(1)`
Áp dụng `(1)` ta có:
`A=1/{2\sqrt{1}+1\sqrt{2}}+1/{3\sqrt{2}+2\sqrt{3}}+...+1/{2005\sqrt{2004}+2004\sqrt{2005}}`
`=1/\sqrt{1}-1/\sqrt{2}+1/\sqrt{2}-1/\sqrt{3}+...+1/\sqrt{2004}-1/\sqrt{2005}`
`=1-1/\sqrt{2005}=1-\sqrt{2005}/{2005}`
Vậy `A=1-\sqrt{2005}/{2005}`
