计算:1/a(a+1)+1/(a+1)(a+2)+

6个回答

1/a(a+1) + 1/(a+1)(a+2) + 1/(a+2)(a+3).1/(a+2003)(a+2004)
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3).+1/(a+2003)-1/(a+2004)
=1/a-1/(a+2004)
=2004/a(a+2004)
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计算 1/[a(a+1)]+1/[(a+1)(a+2)]+……+1/[(a+2003)(a+2004)]
若|a1-1|+(a2-2)2+|a3-3|+(a4-4)4+...+|a2003-2003|+(a2004-2004)
a1,a2,...a2004都是正数,如果M=（a1+a2+...+a2003）·（a2+a3+...+a2004),N
如果A#B=A*A-B*B,那么1-1#2-2#3-...-2003#2004=?
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