已知数列an满足an+1=-1/2an+3/2,a1=4,sn是前n项和,满足|sn-n-2|

Question

已知数列an满足 an+1=-1/2an+3/2,a1=4,sn是前n项和,满足|sn-n-2|

Answer 1

a(n+1)=(-1/2)an+ 3/2
a(n+1) -1=(-1/2)an +1/2=(-1/2)(an -1)
[a(n+1)-1]/(an -1)=-1/2,为定值
a1-1=4-1=3,数列{an -1}是以3为首项,-1/2为公比的等比数列
an -1=3×(-1/2)^(n-1)
an=3×(-1/2)^(n-1) +1
Sn=a1+a2+...+an
=3×[1+(-1/2)+...+(-1/2)^(n-1)] +n
=3×1×[1-(-1/2)ⁿ]/[1-(-1/2)] +n
=2- 2×(-1/2)ⁿ +n
=n+2 +(-1/2)^(n-1)
|Sn-n-2|
=|n+2+(-1/2)^(n-1)-n-2|
=|(-1/2)^(n-1)|
=(1/2)^(n-1)
|Sn-n-2|