计算 1/(x+1)(x+2) + 1/(x+2)(x+3) +……+ 1/(x+2010)(x+2011)

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计算 1/(x+1)(x+2) + 1/(x+2)(x+3) +……+ 1/(x+2010)(x+2011)
计算 1/(x+1)(x+2) + 1/(x+2)(x+3) +……+ 1/(x+2010)(x+2011)

计算 1/(x+1)(x+2) + 1/(x+2)(x+3) +……+ 1/(x+2010)(x+2011)
解: 1/(x+1)(x+2) + 1/(x+2)(x+3) +……+ 1/(x+2010)(x+2011)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+.+1/(x+2010)-1/(x+2011)
=1/(x+1)-1/(x+2011)
=(x+2011-x-1)/[(x+1)(x+2011)]
=2010/[(x+1)(x+2011)]
祝你开心

1/(x+1)(x+2) + 1/(x+2)(x+3) +……+ 1/(x+2010)(x+2011)
=1/(x+1)-1/(x+2) + 1/(x+2)-1/(x+3) +……+ 1/(x+2010)-1/(x+2011)
=1/(x+1)-1/(x+2011)

1/(x+1)(x+2) = 1/(x+1)- 1/(x+2)
1/(x+2)(x+3) = 1/(x+2)- 1/(x+3)
......
正负项相抵消
最后=1/(x+1)- 1/(x+2011) = 2010/(x+1)(x+2011)

先看第一项1/(X+1)(X+2)=1/(X+1) -1/(X+2)
后面同理 1/(X+2)(X+3)=1/(X+2)-1/(X+3)
所以 原式= 1/(x+1)-1/(x+2011)=2010/(x+1)(x+2011)

原式=(1/(x+1)-1/(x+2))+(1/(x+2)-1/(x-3))+…………+1/(x+2010)-1/(x+2011)=2010/(x+1)(x+2011)

原式=(1/(x+1) -1/(x+2))+(1/x+2-1/x+3)+.....+(1/x+2010-1/2011)
=1/(x+1) -1/(x+2011)
把没一项拆成2个分式的差 最后中间的刚好抵消