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電子課本網(wǎng) › 第115頁

第115頁

信息發(fā)布者：

解：?$(1)\sqrt {5+\frac {5}{24}}=\sqrt {\frac {125}{24}}=5\sqrt {\frac {5}{24}}$?

?$(2)$?猜想：?$\sqrt {n+\frac n{n^2-1}}=n\sqrt {\frac n{n^2-1}}(n$?為正整數(shù))

證明：?$\sqrt {n+\frac n{n^2-1}}=\sqrt {\frac {n^3}{n^2-1}}=n\sqrt {\frac n{n^2-1}}，$?等式成立

解：?$\sqrt 3-\sqrt 2＜\sqrt 2-1，$??$\sqrt 4-\sqrt 3＜\sqrt 3-\sqrt 2，$??$\sqrt 5-\sqrt 4＜\sqrt 4-\sqrt 3$?

猜想：?$\sqrt {n+1}-\sqrt n＜\sqrt n-\sqrt {n-1}(n$?是大于等于?$1$?的正整數(shù))

證明：?$\sqrt {n+1}-\sqrt {n}=\frac {(\sqrt {n+1}-\sqrt n)(\sqrt {n+1}+\sqrt n)}{\sqrt {n+1}+\sqrt n}=\frac 1{\sqrt {n+1}+\sqrt n}$?

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

∵?$\sqrt {n+1}+\sqrt n＞\sqrt n+\sqrt {n-1}$?

∴?$\frac 1{\sqrt {n+1}+\sqrt n}＜\frac 1{\sqrt n+\sqrt {n-1}}$?

∴?$\sqrt {n+1}-\sqrt n＜\sqrt n-\sqrt {n-1}$?

解：當(dāng)?$a_{1}=1$?時(shí)，?$a_{2}=\frac 1{1+1}=\frac 12，$??$a_{3}=\frac 1{1+2}=\frac 13···a_n=\frac 1{n}$?

∴?$a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+···+a_{1999}a_{2000}$?

?$=1×\frac 12+\frac 12×\frac 13+\frac 13×\frac 14+···+\frac {1}{1999}×\frac {1}{2000}$?

?$=1-\frac 12+\frac 12-\frac 13+\frac 13-\frac 14+···+\frac {1}{1999}-\frac {1}{2000}$?

?$=1-\frac {1}{2000}$?

?$=\frac {1999}{2000}$?