用裂项法求值1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)

来源:学生作业帮助网 编辑:作业帮 时间:2024/06/17 05:09:39
用裂项法求值1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
x){>eŋM/|y곍MOVU+h+ mC} Ua z԰ ,m $4AaM1P)B#M"},VIv6t!.hWP!oDmN fc f#6 M@!p.@hJ#$'ekd?<;hـ9p'\5 jf+b;Dϓ/5XɞY<=PٌϺZ-dGY-4"9

用裂项法求值1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
用裂项法求值
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)

用裂项法求值1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
=[和1/n(n+1)-和1/(n+1)(n+2)]/2
=[1-1/(n+1)-1/2+1/(n+2)]/2
=1/4-1/2(n+1)(n+2)
1/n(n+2)=[1/n-1/(n+2)]/2
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-1/2(n+1)-1/2(n+2)

1/n(n+2)=1/2(1/n- 1/(n+2))

1/n(n+1)(n+2) =1/2(1/n(n+1)-1/(n+1)(n+2))
不要误会,我可不是抄楼上的!