Question

已知数列｛an｝的前n项和为Tn,且Tn=-an+1／2,n∈N,

标题补充:设bn+2=3㏒1/2an(n∈N),数列｛Cn｝满足Cn=an×bn
求数列｛bn｝的通项公式
求｛Cn｝的前n项和Sn

Answer 1

a1=T1=-a1+1/2，得a1=1/4，
an=Tn-T(n-1)=a(n-1)-an
得an=a(n-1)/2=…=a1/2^(n-1)=1/2^(n+1)
故bn=3(n+1)-2=3n-1

追答

cn=(3n-1)/2^(n+1)
2Sn-Sn=2/2+5/2²+…+(3n-1)/2^n-(2/2²+5/2³+…+(3n-1)/2^(n+1))
Sn=1+(3/2²+3/2³+…+3/2^n)-(3n-1)/2^(n+1)
=1+(3/2)(1-1/2^n)-(3n-1)/2^(n+1)
=5/2-(3n-4)/2^(n+1)