Question

1.求数列1, 3/2^2, 4/2^3,……n/2^n-1,n+1/2^n 的前n项和Sn

1.求数列1, 3/2^2, 4/2^3,……n/2^n-1,n+1/2^n 的前n项和Sn 2.若数列{an}的前n项和为Sn且满足Sn=3/2 an-3，则数列的前n项和Sn 求详解

Answer 1

简单分析一下，详情如图所示

Answer 2

1.
Sn=1+3/2^2+4/2^3+……+n/2^(n-1)+(n+1)/2^n
即Sn=2/2+3/2^2+4/2^3+……+n/2^(n-1)+(n+1)/2^n
1/2
Sn=
2/2^2+3/2^3+……………………+n/2^n+(n+1)/2^(n+1)
两式相减得：
1/2
Sn=2/2+1/2^2+1/2^3+……………………+1/2^n-(n+1)/2^(n+1)
即1/2
Sn=1+1/2^2+1/2^3+……………………+1/2^n-(n+1)/2^(n+1)
1/2
Sn=1+(1/4)*(1-1/2^(n-1))/(1-1/2)
-(n+1)/2^(n+1)
1/2
Sn=3/2-1/2^n-(n+1)/2^(n+1)
Sn=3-(n+3)/2^n.
2.
因Sn=（3/2）an-3①
故n>=2时
S(n-1)=（3/2）a(n-1)-3②
①-②得：
Sn-S(n-1)=（3/2）[an-a(n-1)]
又Sn-S(n-1）=an
故an=（3/2）[an-a(n-1)]
即an=3a(n-1)
当n=1时S1=（3/2）a1-3=a1
解之a1=6
故an是等比数列
an=6*3^(n-1)=2*3^n
所以Sn=（3/2）an-3
=（3/2）*2*3^n-3
=3^(n+1)-3.