1／1×3×5＋1／3×5×7＋1／5×7×9＋···＋1／2001×2003×2005等于多少？（列方程解）

解：1／[(2k-1)(2k+1)(2k+3)]={1／[(2k-1)(2k+1)] - 1／[(2k+1)(2k+3)]}／4,
将k=1,2,3,·······,2001代入，
原式=[1/(1×3)-1/(3×5)+1/(3×5)-1/(5×7)+·····+1/(2001×2003)-1/(2003×2005)]/4
=1/12-1/(4×2003×2005)。
公式：(∑是求和的记号）∑1/(2n-1)(2n+1)(2n+3)=1/4 - 1/[4(2n+1)(2n+3)] .

2/[n(n+1)(n+2)]=1/[(n(n+1)]-1/[(n+1)(n+2)]=1/n-1/(n+1)+1/(n+1)-1/(n+2)=1/n-1/(n+2)
2／(1·3·5)＋2／(3·5·7)＋2／(5·7·9)＋···＋2／(2001·2003·2005)
=(1-1/5)+(1/3-1/7)+(1/5-1/9)+(1/7-1/11)+···+(1/1995-1/1999)+(1/1997-1/2001)+(1/1999-1/2003)+(1/2001-1/2005)
=1+1/3+1/2003-1/2005=4/3+2/(2003·2005)
原式=2/3+1/(2003·2005)=8032033/12048045