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](/uploads/image/z/8893055-47-5.jpg?t=1%2Fn%28n%2B1%29%2B1%2F%28n%2B1%29%28n%2B2%29%2B1%2F%28n%2B2%29%28n%2B3%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28n%2B2005%29%28n%2B2006%29%E6%B1%82%E5%87%BA%E5%BD%93n%3D1%E6%97%B6%2C%E8%AF%A5%E4%BB%A3%E6%95%B0%E5%BC%8F%E7%9A%84%E5%80%BC)
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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