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

Question

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

Answer 1

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)] .