Question

在数列an中,a1=1.an+1=（1+1/n)an +（n+1)/2^n (1)设bn=an/n,求数列｛bn｝的通项

Answer 1

(1)

a(n+1)=(1+ 1/n)an+(n+1)/2ⁿ=[(n+1)/n]an+(n+1)/2ⁿ

a(n+1)/(n+1)=an/n +1/2ⁿ

a(n+1)/(n+1)-an/n=1/2ⁿ

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

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

…………

a2/2-a1/1=1/2

累加

an/n -a1/1=1/2+1/2²+...+1/2^(n-1)=(1/2)[1-(1/2)^(n-1)]/(1-1/2)=1-1/2^(n-1)

an/n=a1+1-1/2^(n-1)=2- 1/2^(n-1)

bn=an/n bn=2-1/2^(n-1)

数列{bn}的通项公式为bn=2- 1/2^(n-1)

(2)

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

Sn=a1+a2+...+an=2(1+2+...+n) -[1/1+2/2+3/2²+...+n/2^(n-1)]

令Cn=1/1+2/2+3/2²+...+n/2^(n-1)

则Cn/2=1/2+2/2²+...+(n-1)/2^(n-1)+n/2ⁿ

Cn-Cn/2=Cn/2=1+1/2+1/2²+...+1/2^(n-1)-n/2ⁿ

=1×(1-1/2ⁿ)/(1-1/2)-n/2ⁿ

=2-(n+2)/2ⁿ

Cn=4 -(n+2)/2^(n-1)

Sn=2(1+2+...+n) -Cn

=2n(n+1)/2 -4+(n+2)/2^(n-1)

=(n+2)/2^(n-1) +n²+n -4