观察下列关系式:1/(x-1)(x-2)=1/(x-2)-1/(x-1); 1/(x-2)(x-3)=1/(x-3)-1
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观察下列关系式:1/(x-1)(x-2)=1/(x-2)-1/(x-1); 1/(x-2)(x-3)=1/(x-3)-1/(x-2);
观察下列关系式:1/(x-1)(x-2)=1/(x-2)-1/(x-1);
1/(x-2)(x-3)=1/(x-3)-1/(x-2);
1/(x-3)(x-4)=1/(x-4)-1/(x-3);
1/(x-4)(x-5)=1/(x-5)-1/(x-4);
.
(1)你可以归纳出一般结论是____________.
(2)请利用上述结论计算:1/(x+1)+1/(x-1)(x+2)+1/(x-2)(x-3)+...+1/(x-99)(x-100).
观察下列关系式:1/(x-1)(x-2)=1/(x-2)-1/(x-1);
1/(x-2)(x-3)=1/(x-3)-1/(x-2);
1/(x-3)(x-4)=1/(x-4)-1/(x-3);
1/(x-4)(x-5)=1/(x-5)-1/(x-4);
.
(1)你可以归纳出一般结论是____________.
(2)请利用上述结论计算:1/(x+1)+1/(x-1)(x+2)+1/(x-2)(x-3)+...+1/(x-99)(x-100).
可以归纳出一般的结论是:相邻的两个自然数积的倒数等于这两个自然数的倒数之差!
1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)
=1/(x-1)-1/(x-2)+1/(x-2)-1/(x-3)+……+1/(x-99)-1/(x-100)
=1/(x-1)-1/(x-100)
=[(x-100)-(x-1)]/[(x-1)(x-100)]
=-99/[(x-1)(x-100)]
1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)
=1/(x-1)-1/(x-2)+1/(x-2)-1/(x-3)+……+1/(x-99)-1/(x-100)
=1/(x-1)-1/(x-100)
=[(x-100)-(x-1)]/[(x-1)(x-100)]
=-99/[(x-1)(x-100)]
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