设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知bn>0(n∈N*),a1=b1=1,a2+b3=a3,S5=5(T3+b2)1）求数列{an}和{bn}的通项公式2）求和：b1/T1T2+b1/T2T3+…+bn/TnTn+1

Question

设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知bn>0(n∈N*),a1=b1=1,a2+b3=a3,S5=5(T3+b2)
1）求数列{an}和{bn}的通项公式
2）求和：b1/T1T2+b1/T2T3+…+bn/TnTn+1

Answer 1

1、
设等差数列{an}公差为d,等比数列{bn}公比为q.
a2+b3=a3 b3=d b1q²=d b1=1代入,得d=q²
S5=5(T3+b2) 5a1+10d=5(b1+b1q+b1q²+b1q)
a1=1 b1=1 d=q²代入,整理,得
q²-2q=0
q(q-2)=0
q=0(等比数列公比不等于0,舍去)或q=2
d=q²=2²=4
an=a1+(n-1)d=1+4(n-1)=4n-3
n=1时,a1=4-3=1,同样满足.
bn=b1q^(n-1)=2^(n-1)
n=1时,b1=2^0=1,同样满足.
数列{an}的通项公式为an=4n-3；数列{bn}的通项公式为bn=2^(n-1).
2、
Tn=b1(q^n-1)/(q-1)=2^n-1 Tn+1=2^(n+1) -1
1/Tn-1/T(n+1)=1/(2^n-1)-1/[2^(n+1)-1]
=[2^(n+1)-1-2^n+1]/[(2^n-1)(2^(n+1)-1)]
=2^n/[(2^n-1)(2^(n+1)-1)]
=2×bn/(TnTn+1)
bn/(TnTn+1)=(1/2)[1/Tn-1/T(n+1)]
b1/(T1T2)+b2/(T2T3)+...+bn/[TnT(n+1)]
=(1/2)[1/T1-1/T2+1/T2-1/T3+...+1/Tn-1/T(n+1)]
=(1/2)[1/T1-1/T(n+1)]
=(1/2)[1-1/[2^(n+1)-1]]
=(1/2)[2^(n+1)-2]/[2^(n+1)-1]
=(2^n-1)/[2^(n+1)-1]