Question

设数列｛An｝前n项和sn=4-an-(1/2^n-2),求An?

Answer 1

a1=s1=4-a1-1/2^(1-2)=4-a1-2=2-a1
a1=1
sn=4-an-1/2^(n-2)
s（n-1）=4-a（n-1）-1/2^（n-3）
sn=s（n-1）+an
4-an-1/2^(n-2)=4-a（n-1）-1/2^（n-3）+an
2an-a（n-1）+1/2^(n-2)-1/2^（n-3）=0
2an-a（n-1）+1/2^(n-3)=0
设2[an+x]-[a(n-1)+x]=0
x=1/2^(n-3)
2[an+1/2^(n-3)]-[a(n-1)+1/2^(n-3)]=0
2[an+1/2^(n-3)]=[a(n-1)+1/2^(n-3)]
设an+1/2^(n-3)=bn,
b1=a1+1/2^(1-3)=1+4=5
2bn=b(n-1)
bn=5(1/2)^(n-1)
an=bn-1/2^(n-3)
=5(1/2)^(n-1)-1/2^(n-3)
=5(1/2)^(n-1)-4(1/2)^(n-1)
=(1/2)^(n-1)