第32页 - 启东中学作业本八年级数学人教版 - 05网零5网 0五网新知语文网

$\frac{1}{a} $

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$\frac{x+6}{x+7}-\frac{x+5}{x+6}=\frac{1}{x+6}-\frac{1}{x+7}\ $

$\frac {x+n+1}{x+n+2}-\frac {x+n}{x+n+1}=\frac {1}{x+n+1}-\frac {1}{x+n+2}$

$(3)证明:\frac {x+n+1}{x+n+2}-\frac {x+n}{x+n+1}=\frac{(x+n+2)-1}{x+n+2}-\frac{(x+n+1)-1}{x+n+1}$

$=(1-\frac{1}{x+n+2})-(1-\frac{1}{x+n+1}) =\frac{1}{x+n+1}-\frac{1}{x+n+2},即\frac{x+n+1}{x+n+2}-\frac {x+n}{x+n+1}=\frac {1}{x+n+1}-\frac {1}{x+n+2}$

$解:∵S_{1}=\frac{1}{a}$

$∴S_{2}=-S_{1}-1=-\frac{1}{a}-1=-\frac{a+1}{a}$

$∴S_{3}=\frac{1}{S_{2}}=-\frac{a}{a+1}$

$解:∵S_{1}=\frac{1}{a},∴S_{2}=-\frac{a+1}{a}$

$∴S_{3}=-\frac{a}{a+1}$

$∴S_{4}=-\frac{1}{a+1},∴S_{5}=-(a+1)\ $

$∴S_{6}=a,∴S_{7}=\frac{1}{a}···\ $

$∴S_{1}+S_{2}+S_{3}+S_{4}+S_{5}+S_{6}$

$=\frac{1}{a}+(-\frac{a+1}{a})+(-\frac{a}{a+1})+(-\frac{1}{a+1})+[-(a+1)]+a=-3\ $

$∵2022÷6=337$

$∴S_{1}+S_{2}+S_{3}+···+S_{2022}$

$=(-3)×337=-1011 $