设数列{an}满足a1+3a2+3^2a3+3^3a4+······+3^n-1×an=3分之n,n属于N* 【1】求数列,{an}的通项公式【2】设b=a的n次幂分之n,求数列{bn}的前n项和sn
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设数列{an}满足a1+3a2+3^2a3+3^3a4+······+3^n-1×an=3分之n,n属于N*【1】求数列,{an}的通项公式【2】设b=a的n次幂分之n,求数列{bn}的前n项和sn设
设数列{an}满足a1+3a2+3^2a3+3^3a4+······+3^n-1×an=3分之n,n属于N* 【1】求数列,{an}的通项公式【2】设b=a的n次幂分之n,求数列{bn}的前n项和sn
设数列{an}满足a1+3a2+3^2a3+3^3a4+······+3^n-1×an=3分之n,n属于N* 【1】求数列,{an}的通项公式
【2】设b=a的n次幂分之n,求数列{bn}的前n项和sn
设数列{an}满足a1+3a2+3^2a3+3^3a4+······+3^n-1×an=3分之n,n属于N* 【1】求数列,{an}的通项公式【2】设b=a的n次幂分之n,求数列{bn}的前n项和sn
1、a1+3a2+3^2a3+3^3a4+······+3^n-1×an=3分之n
a1+3a2+3^2a3+3^3a4+······+3^n-2×a(n-1)=3分之n-1
两式相减3^n-1×an=1/3
即an=1/3^n
2、bn=n/a^n
sn=1/a+2/a^2+3/a^3+……+n/a^n
(1/a)sn=1/a^2+2/a^3+……+n/a^(n+1)
(1- 1/a) sn=1/a+1/a^2+1/a^3+……+1/a^n-n/a^(n+1)
(1- 1/a) sn=(1+1/a^n)/(a-1)-n/a^(n+1)
sn=.
这是乘公比错位相减法,经常用的
best wish
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