電子課本網(wǎng) › 第115頁

第115頁

信息發(fā)布者：

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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