1⼀3+3⼀(3^2)+5⼀(3^3)+...+(2n-1)⼀(3^n)=? 急求!用错位相减做啊!明早就要!老师说做错抄10遍…

2024-12-18 16:11:56
推荐回答(4个)
回答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)

回答2:

asd

回答3:

去找人啊

回答4:

活该