第115页 - 苏科版数学补充习题九年级上下册答案

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

$(2)猜想：\sqrt {n+\frac n{n^2-1}}=n\sqrt {\frac n{n^2-1}}(n为正整数)$

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

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

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

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

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

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

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

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

$解：当a_{1}=1时，a_{2}=\frac 1{1+1}=\frac 12，a_{3}=\frac 1{1+2}=\frac 13···a_n=\frac 1{n}$

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

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

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

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

$=\frac {1999}{2000}$