Question

数列an的前n项和记为sn,且sn=3/2(an-1)bn满足bn+1=1/4bn,且b1=1

求an与bn通项公式
cn满足cn=anlog2bn,其前n项和为tn，求tn

Answer 1

当n=1有s1=a1=3/2(a1-1)

所以a1=3

n>1时有：
an=Sn-Sn-1 = 3/2 (an -1) - 3/2 (a(n-1) - 1）

an=3*a(n-1)
所以an = 3^n
又因为bn+1=1/4bn
且b1=1

所有bn=(1/4)^(n-1)
Cn=anlog2bn=3^nlog2(1/4)^(n-1)=3^n(-2n+2)
Tn=C1+C2+……+Cn=0+3^2X(-2)+3^3X(-4)+…… +3^n(-2n+2)
3Tn= 3^3X(-2)+3^4X(-4)+……+3^n(-2n) +3^(n+1)(-2n+2)
2Tn= 3^(n+1)(-2n+2)-2(3^3+3^4+……+3^n)
Tn= 3^(n+1)(-n+1)-(3^3+3^4+……+3^n)=3^(n+1)(-n+1)-(3^n-1)/2+3+9
=3^(n+1)(-n+1)-(3^n-1)/2+12