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+2
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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)=?
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)
∴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)
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