求Sn=1 (1/2)+3 (1/2)²+5 (1/2)³+...+(2n-1) (1/2)n次方 计算

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/05 19:56:27
求Sn=1 (1/2)+3 (1/2)²+5 (1/2)³+...+(2n-1) (1/2)n次方 计算
x){)8PC}{:l#kmSTckm===m <]CM f޳5 M۩b$`;qjʆ(Ƃ,1͍djS04\$CaC5R#OPl]O?mxgm@J(HBk?jXDPn^.AiiA 6ԅ9x逨5̓:FAHCٺFV4HiAuld1DX 12Q@

求Sn=1 (1/2)+3 (1/2)²+5 (1/2)³+...+(2n-1) (1/2)n次方 计算
求Sn=1 (1/2)+3 (1/2)²+5 (1/2)³+...+(2n-1) (1/2)n次方 计算

求Sn=1 (1/2)+3 (1/2)²+5 (1/2)³+...+(2n-1) (1/2)n次方 计算
Sn=1 (1/2)+3 (1/2)^²+5 (1/2)^³+...+(2n-1) (1/2)^n
1/2*Sn= 1 (1/2)^²+3 (1/2)^³+...+(2n-3) (1/2)n+(2n-1) (1/2)^(n+1)
上式减下式:
1/2*Sn=1/2+2[(1/2)^2+(1/2)^3+……+(1/2)^n]-(2n-1) (1/2)^(n+1)
=1/2-(2n-1) (1/2)^(n+1)+2*(1/2)^2[1-(1/2)^(n-1)]/(1/2)
=1/2-(2n-1) (1/2)^(n+1)+[1-(1/2)^(n-1)]
=3/2-n(1/2)^n +1/2(1/2)^n-2(1/2)^n
=3/2-(n+3/2)/2^n
∴Sn=3-(2n+3)/2^n