已知数列{an},{bn},{an}为公比q>0的等比数列,Sn,Tn分别为{an},{bn}的前n项和且S4=20,S8=340,b1=1,bn=3/4nSn,求Tn

已知数列{an},{bn},{an}为公比q>0的等比数列,Sn,Tn分别为{an},{bn}的前n项和且S4=20,S8=340,b1=1,bn=3/4nSn,求Tn
已知数列{an},{bn},{an}为公比q>0的等比数列,Sn,Tn分别为{an},{bn}的前n项和
且S4=20,S8=340,b1=1,bn=3/4nSn,求Tn

已知数列{an},{bn},{an}为公比q>0的等比数列,Sn,Tn分别为{an},{bn}的前n项和且S4=20,S8=340,b1=1,bn=3/4nSn,求Tn
S8/S4=[a1(q^8-1)/(q-1)]/[a1(q^4-1)/(q-1)]
=[a1(q^4+1)(q^4-1)/(q-1)]/[a1(q^4-1)/(q-1)]
=q^4+1=340/20=17
q^4=16
q>0 q=2
S4=a1(q^4-1)/(q-1)=a1(16-1)/(2-1)=15a1=20
a1=4/3
Sn=a1(q^n-1)/(q-1)=(4/3)(2^n-1)/(2-1)=(4/3)(2^n-1)
bn=(3/4)nSn=(3/4)(4/3)n(2^n-1)=n(2^n-1)
Tn=b1+b2+...+bn
=1×(2^1-1)+2×(2^2-1)+...+n(2^n-1)
=(1×2^1+2×2^2+...+n×2^n)-(1+2+...+n)
令Cn=1×2^1+2×2^2+...+n×2^n
则2Cn=1×2^2+2×2^3+...+(n-1)×2^n+n×2^(n+1)
Cn-2Cn=-Cn=2^1+2^2+...+2^n-n×2^(n+1)
=2(2^n-1)/(2-1)-n×2^(n+1)
=(1-n)×2^(n+1)-2
Cn=(n-1)×2^(n+1)+2
Tn=(1×2^1+2×2^2+...+n×2^n)-(1+2+...+n)
=(n-1)×2^(n+1)+2-n(n+1)/2