cn=(n-1)/3^n ,求cn的前n项和TN

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 01:18:57
cn=(n-1)/3^n ,求cn的前n项和TN

cn=(n-1)/3^n ,求cn的前n项和TN
cn=(n-1)/3^n ,求cn的前n项和TN

cn=(n-1)/3^n ,求cn的前n项和TN
cn=(n-1)*(1/3)^n
等差乘等比类型求和利用错项相减法:
Tn=c1+c2+...+c(n-1)+cn
即:
Tn=(1-1)*(1/3)^1+(2-1)*(1/3)^2+...+(n-1-1)*(1/3)^(n-1)+(n-1)*(1/3)^n ----(1)
左右同时乘以公比(1/3)得:
(1/3)Tn= (1-1)*(1/3)^2+(2-1)*(1/3)^3+...+(n-2)*(1/3)^n+(n-1)*(1/3)^(n+1) ----(2)
(1)-(2)得:
(2/3)Tn=(1-1)*(1/3)^1+[(1/3)^2+(1/3)^3+...+(1/3)^n]-(n-1)*(1/3)^(n+1)
(2/3)Tn=0+(1/9)*[1-(1/3)^(n-1)]/(1-1/3)-[(n-1)/3]*(1/3)^n
(2/3)Tn=(1/6)-(1/6)*(1/3)^(n-1)-[(n-1)/3]*(1/3)^n
(2/3)Tn=(1/6)-[1/2+(n-1)/3]*(1/3)^n
则:
Tn=(1/4)-[(2n+1)/4]*(1/3)^n