高中数列裂项相消

2025-06-28 20:44:01

推荐回答（2个）

回答1：

求和：S‹n›=1/(1×2×3)+1/(2×3×4)+1/(3×4×5)+.......+1/n(n+1)(n+2)
解：通项a‹n›=1/n(n+1)(n+2)=(1/2)[1/n(n+1)-1/(n+1)(n+2)]=(1/2){[(1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}
故S‹n›=(1/2){[(1-1/2)-(1/2-1/3)]+[(1/2-1/3)-(1/3-1/4)]+[(1/3-1/4)-(1/4-1/5)]+...+[(1/n-1/(n+1)]-
-[1/(n+1)-1/(n+2)]}
=(1/2){(1-1/2)-[1/(n+1)-1/(n+2)]}=(1/2)[1/2-1/(n+1)(n+2)]

回答2：

解：令an=（1/2）*[1/n(n+1)-1/(n+1)(n+2)]
所以Sn=（1/2）*[1/2-\1/6+1/6-1/12+.......+1/n(n+1)-1/(n+1)(n+2)]
=（1/2）*[1/2-1/(n+1)(n+2)]