Question

已知数列{an}的前n项和Sn=-an-(1/2)^(n-1)+2(n为正整数).令bn=2^n*an,求证数列{bn}

Answer 1

Sn=-an-(1/2)^(n-1)+2

所以 S(n-1)=-a(n-1)-(1/2)^(n-2)+2

相减

Sn-S(n-1)=an=-an-(1/2)^(n-1)+a(n-1)+(1/2)^(n-2)

(1/2)^(n-2)-(1/2)^(n-1)=(1/2)^(n-2)-1/2*(1/2)^(n-2)=(1/2)^(n-2)

2an=a(n-1)+(1/2)^(n-2)

2an-(1/2)^(n-3)=a(n-1)+(1/2)^(n-2)-(1/2)^(n-3)

2[an-(1/2)^(n-2)]=a(n-1)-(1/2)^(n-3)

[an-(1/2)^(n-2)]/[a(n-1)-(1/2)^(n-3)]=1/2

所以an-(1/2)^(n-2)是等比数列,q=1/2

a1=S1=-a1-(1/2)^(1-1)+2=1-a1

a1=1/2

所以a1-(1/2)^(1-2)=1/2-2=-3/2

所以an-(1/2)^(n-2)=（-3/2）*(1/2)^(n-1)=(-3/4)*(1/2)^(n-2)

所以an=(-3/4)*(1/2)^(n-2)+(1/2)^(n-2)

即an=(1/4)*(1/2)^(n-2)=(1/2)^n

bn=2^n*an=1,是常数列

所以bn是等差数列