数列求和 已知an=1/[n(n+3)],求Sn=

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/25 09:43:10
数列求和已知an=1/[n(n+3)],求Sn=数列求和已知an=1/[n(n+3)],求Sn=数列求和已知an=1/[n(n+3)],求Sn=用裂项相消法.因为3an=3/[n(n+3)]=1/n-

数列求和 已知an=1/[n(n+3)],求Sn=
数列求和 已知an=1/[n(n+3)],求Sn=

数列求和 已知an=1/[n(n+3)],求Sn=
用裂项相消法.
因为 3an=3/[n(n+3)]=1/n -1/(n+3)
所以3Sn=3a1+3a2+3a3+3a4+...+3an
=(1-1/4) +(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+...+1/n -1/(n+3)
=1+1/2+1/3 -1/(n+1) -1/(n+2)-1/(n+3)
=11/6 +(3n²+12n+11)/[(n+1)(n+2)(n+3)]
然后,两边除以3,整理即可.

sn=1/3(1-1/4+1/2-1/5+1/3-1/6+1/4-1/7+........+1/n-1/(n+3)=11/18-(3n^2+12+8)/3(n+1)(n+2)(n+3)可以写的再详细点吗1/3(1-1/4+1/2-1/5+1/3-1/6+1/4-1/7+1/5-1/8+.........................+1/(n-3)-1/n+1/(n-2)-1/(n+1)+1/n...

全部展开

sn=1/3(1-1/4+1/2-1/5+1/3-1/6+1/4-1/7+........+1/n-1/(n+3)=11/18-(3n^2+12+8)/3(n+1)(n+2)(n+3)

收起