第1个回答 2010-03-27
1.
An=2a(n-1)+2^n
An/2^n=2a(n-1)/2^n+1
An/2^n=a(n-1)/2^(n-1)+1
An/2^n即首项为A1/2^1=1/2,公差为1的等差数列。
2.
由上得:
An/2^n=a(n-1)/2^(n-1)+1
An/2^n=(1/2)+1*(n-1)=n-1/2
An=n*2^n-2^(n-1)
Sn=[1*2^1+2*2^2+3*2^3+……+(n-2)*2^(n-2)+(n-1)*2^(n-1)+n*2^n]-[2^n-1]
2Sn=[1*2^2+2*2^3+3*2^4+……+(n-2)*2^(n-1)+(n-1)*2^n+n*2^(n+1)]-[2^(n+1)-2]
两式相减:
-Sn=[2^1+2^2+2^3+2^4+……+2^(n-1)+2^n-n*2^(n+1)]-[-2^n+1]
=2(2^n-1)-n*2^(n+1)+2^n-1
=(2-2n+1)2^n-3
=(-2n+3)2^n-3
Sn=(2n-3)2^n+3