cn=(n-1)*(1/3)^n
等差乘等比類(lèi)型求和利用錯(cuò)項(xiàng)相減法：
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)
左右同時(shí)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
則：
Tn=(1/4)-[(2n+1)/4]*(1/3)^n