Question

1/3+3/(3^2)+5/(3^3)+...+(2n-1)/(3^n)=? 急求！用错位相减做啊！明早就要！老师说做错抄10遍…

Answer 1

两边同乘3
3Sn=1+3*(1/3)+5*(1/3)^2+...+(2n-1)*(1/3)^(n-1)
Sn = 1*(1/3)+3*(1/3)^2+...+(2n-3)*(1/3)^(n-1)+(2n-1)*(1/3)^n
两式相减
2Sn=1+2*(1/3)+2*(1/3)^2+...+2*(1/3)^(n-1)-(2n-1)*(1/3)^n
2Sn=1-(2n-1)*(1/3)^n+2*[1/3+(1/3)^2+...+(1/3)^(n-1)]
方括号内是一等比数列求和，首项1/3,公比1/3,一共(n-1)项（看指数1,2,...,n-1）
2Sn=1-(2n-1)*(1/3)^n+2*(1/3)(1-(1/3)^(n-1))/(1-1/3)
2Sn=1-(2n-1)*(1/3)^n+1-(1/3)^(n-1)
=2-(2n-1+3)*(1/3)^(n)
=2-(2n+2)*(1/3)^(n)
Sn=1-(n+1)*(1/3)^(n)

Answer 2

asd

Answer 3

去找人啊

Answer 4

活该