求和1×(1/3)+3×(1/9)+5×(1/27)+…+(2n－1) ×(1/3^n)
热心网友
分析：这是一道等差数列乘以等比数列求和题 方法用 错位相减法 先写出Sn 再两边同乘以等比数列的公比 q Sn = 1*(1/3) + 3*(1/3)^2 + 5*(1/3)^3 + … + (2n-1)*(1/3)^n (1/3)Sn = 1*(1/3)^2 + 3*(1/3)^3 + 5*(1/3)^4 + … + (2n-1)*(1/3)^(n+1)用一式减二式： (2/3)Sn = 1/3 + 2*[(1/3)^2+(1/3)^3+…+(1/3)^n] - (2n-1)*(1/3)^(n+1) (2/3)Sn = 1/3 + 2*[(1/3)^2-(1/3)^(n+1)]/(2/3) - (2n-1)*(1/3)^(n+1) (2/3)Sn = 1/3 + 1/3 - (1/3)^n - (2n-1)*(1/3)^(n+1) Sn = 1/2 + 1/2 - (3/2)(1/3)^n - [(2n-1)/2]*(1/3)^n Sn = 1 - (n+1)*(1/3)^n。
热心网友
设s=1×(1/3)+3×(1/9)+5×(1/27)+…+(2n－1) ×(1/3^n) 则：3s=1×1+3×(1/3)+5×(1/9)+…+(2n－1) ×[1/3^(n-1)]3s-s=2s=1+2×[(1/3)+(1/9)+...+(1/3^(n-1))]-(2n－1) ×(1/3^n) =2×[1+(1/3)+(1/9)+...+(1/3^(n-1))]-1-(2n－1) ×3^(-n) =2×[1-3^(-n)]/(1-1/3)-1-(2n－1) ×3^(-n)=3×[1-3^(-n)]-1-(2n－1) ×3^(-n)=3-3×3^(-n)-1-(2n－1)×3^(-n)=2-(2n+2)×3^(-n)∴s=1-(n+1)×3^(-n)