Question

数列{an}，a1=1,a(n+1)=2an+2^n. (1)设bn=(an/2^(n-1))，证：bn是等差数列； （2）求数列{an}的前n项和Sn

Answer 1

1.
A(n+1)=2An+2^n
两边同除2^n
A(n+1)/2^n=2An/2^n+1
B(n+1)=Bn+1
{Bn}是公差为1的等差数列

2.
B1=A1/2^0=1
Bn=n=An/2^(n-1)
An=n×2^(n-1)
Sn=A1+A2+A3+……+An
=1×2^0+2×2^1+3×2^2+……+n×2^(n-1)
2Sn=1×2^1+2×2^2+3×2^3+……+n×2^n
两式错位相减
Sn-2Sn=1+[(2×2^1-1×2^1)+(3×2^2-2×2^2)+……+n×2^(n-1)-(n-1)×2^(n-1)]-n×2^n
=1+(2^1+2^2+……+2^(n-1))-n×2^n
=1×(1-2^n)/(1-2)-n×2^n
=(1-n)×2^n-1
Sn=(n-1)×2^n+1