(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)
數(shù)列{bn}的通項(xiàng)公式為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)
則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