数列An满足A1=2,An+1-An=3*2^2n-1.〔Ⅰ〕求数列An的通项公式；〔Ⅱ〕令Bn=nAn,求数列Bn的前n项和?

Question

数列An满足A1=2,An+1-An=3*2^2n-1.〔Ⅰ〕求数列An的通项公式；〔Ⅱ〕令Bn=nAn,求数列Bn的前n项和?

Answer 1

An-(An-1)=3*2^2n-3
(An-1)-(An-2)=3*2^2n-5
……
A2-A1=3*2^2
相加得
An-A1=3*2（1-4^n-1）/(1-4)
An-2=2*4^n-1-2
An=2*4^(n-1)=2^(2n-1)
Bn=n*2^(2n-1)
Sn=1*2^1+2*2^3+3*2^5+4*2^5……+n*2^(2n-1)
4Sn= 1*2^3+2*2^5+3*2^5……+(n-1)*2^(2n-1)+n*2^(2n+1)
错位相减得
-3Sn=2+2^3+2^5+2^7+……2^(2n-1)-n*2^(2n+1)
-3Sn=2/3(4^n -1)-)-n*2^(2n+1)
Sn=-2/9(4^n -1)+n/3*2^(2n+1)