一道分式数学题,1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)=?
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![一道分式数学题,1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)=?](/uploads/image/z/2611097-17-7.jpg?t=%E4%B8%80%E9%81%93%E5%88%86%E5%BC%8F%E6%95%B0%E5%AD%A6%E9%A2%98%2C1%2Fa%28a%2B1%29%2B1%2F%28a%2B1%29%28a%2B2%29%2B1%2F%28a%2B2%29%28a%2B3%29%2B.%2B1%2F%28a%2B2004%29%29%28a%2B2005%29%3D%3F)
一道分式数学题,1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)=?
一道分式数学题,
1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)=?
一道分式数学题,1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)=?
∵1/a(a+1)=1/a -1/(a+1)
1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)
∴原式=1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004))(a+2005)
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+..-1/(a+2004)+1/(a+2004)-1/(a+2005)
=1-1/(a+2005)
=(a+2004)/(a+2005)
1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.....+1/(a+2004))(a+2005)
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+..-1/(a+2004)+1/(a+2004)-1/(a+2005)
=1-1/(a+2005)
=(a+2004)/(a+2005)
分数太少了,但是我还是要说明一下方法了,助人为乐嘛!呵呵!用裂项法,原式=1/a-1/(a+1)+...+1/(a+2004)-1/(a+2005)=1/a-1/(a+2005)
给你个公式:1/n(n+1)=1/n-1/(n+1)。所以原式=2005/[a(a+2005)]。