1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+……+1/(n+2005)(n+2006)求出当n=1时,该代数式的值

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1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+……+1/(n+2005)(n+2006)求出当n=1时,该代数式的值
1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+……+1/(n+2005)(n+2006)求出当n=1时,该代数式的值

1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+……+1/(n+2005)(n+2006)求出当n=1时,该代数式的值
原式=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+……+1/(n+2005)-1/(n+2006)
中间正负抵消
=1/n-1/(n+2006)
=1-1/2007
=2006/2007

注：看到这类题，即都是分数加减首先考虑能不能通过拆项互相抵消

考虑到1/n(n+1)=1/n-1/(n+1)
1/(n+1)(n+2)=1/(n+1)-1/(n+2)
……
……
1/(n+2005)(n+2006)=1/(n+2005)-1/(n+2006)
故代入原式得
原式=1/...

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注：看到这类题，即都是分数加减首先考虑能不能通过拆项互相抵消

考虑到1/n(n+1)=1/n-1/(n+1)
1/(n+1)(n+2)=1/(n+1)-1/(n+2)
……
……
1/(n+2005)(n+2006)=1/(n+2005)-1/(n+2006)
故代入原式得
原式=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+……+1/(n+2005)-1/(n+2006)
相邻的加减号相互抵消得
原式=1/n-1/(n+2006)
=1-1/2007
=2006/2007

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