已知数列{an}满足:na1+(n-1)a2……+2an-1+an}=( 2/3)n次+(2/3
已知数列{an}满足:na1+(n-1)a2……+2an-1+an}=( 2/3)n次+(2/3
已知数列{an}满足:na1+(n-1)a2……+2an-1+an}=( 2/3)n次+(2/3)n-1+……+(2/3),数列{an}的前n项和 为sn，设bn=nSn 1、求数列{an}的通项公式 2、求b1+b3+b3+……+bn的值
已知数列{an}满足:na1+(n-1)a2……+2an-1+an}=( 2/3)n次+(2/3)n-1+……+(2/3),数列{an}的前n项和 为sn，设bn=nSn 1、求数列{an}的通项公式 2、求b1+b3+b3+……+bn的值
赞 kmasato 幼苗
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na1+(n-1)a2……+2a(n-1)+an=( 2/3)^n+(2/3)^(n-1)+……+(2/3) (1)
n=1,a1= 2/3
(n-1)a1+(n-2)a2……+2a(n-2)+a(n-1)=( 2/3)^(n-1)+(2/3)^(n-2)+……+(2/3) (2)
(1)-(2)
a1+a2+...+an = (2/3)^n (3)
a1=2/3 satisfy (3)
ie
Sn=a1+a2+...+an = (2/3)^n
let
S=1.(2/3)^1+2.(1/3)^2+.+n(2/3)^n (1)
(2/3)S= 1.(2/3)^2+2.(1/3)^3+.+n(2/3)^(n+1) (2)
(1)-(2)
(1/3)S = [2/3+(2/3)^2+...+(2/3)^n] -n(2/3)^(n+1)
= 2(1- (2/3)^n) -n(2/3)^(n+1)
S = 6(1- (2/3)^n) -n(2/3)^n
= 6 -(n+6).(2/3)^n
bn = nSn
=n(2/3)^n
b1+b2+b3+...+bn = S =6 -(n+6).(2/3)^n
n=1,a1= 2/3
(n-1)a1+(n-2)a2……+2a(n-2)+a(n-1)=( 2/3)^(n-1)+(2/3)^(n-2)+……+(2/3) (2)
(1)-(2)
a1+a2+...+an = (2/3)^n (3)
a1=2/3 satisfy (3)
ie
Sn=a1+a2+...+an = (2/3)^n
let
S=1.(2/3)^1+2.(1/3)^2+.+n(2/3)^n (1)
(2/3)S= 1.(2/3)^2+2.(1/3)^3+.+n(2/3)^(n+1) (2)
(1)-(2)
(1/3)S = [2/3+(2/3)^2+...+(2/3)^n] -n(2/3)^(n+1)
= 2(1- (2/3)^n) -n(2/3)^(n+1)
S = 6(1- (2/3)^n) -n(2/3)^n
= 6 -(n+6).(2/3)^n
bn = nSn
=n(2/3)^n
b1+b2+b3+...+bn = S =6 -(n+6).(2/3)^n
1年前
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