已知数列{an}为等差数列,a3=5,a7=13,数列{}bn的前n项和为Sn,且有Sn=2bn-11)求{an},{bn}的通项公...已知数列{an}为等差数列,a3=5,a7=13,数列{}bn的前n项和为Sn,且有Sn=2bn-11)求{an},{bn}的通项公式2)若Cn=anbn,{Cn}

Question

已知数列{an}为等差数列,a3=5,a7=13,数列{}bn的前n项和为Sn,且有Sn=2bn-1 1)求{an},{bn}的通项公...
已知数列{an}为等差数列,a3=5,a7=13,数列{}bn的前n项和为Sn,且有Sn=2bn-1
1)求{an},{bn}的通项公式
2)若Cn=an bn,{Cn}的前n项和为Tn,求Tn

Answer 1

(1)d=(a7-a3)/(7-3)=8/4=2
∴an=5+(n-3)×2=2n-1
b1=S1=2b1-1
∴b1=1
n≥2时,bn=Sn-Sn-1=
2bn-2bn-1
即2bn-1=bn
bn/bn-1=2
则bn=1×2^(n-1)=2^(n-1)
(2)Cn=(2n-1)×2^(n-1)
2Cn=(2n-1)×2^n
则Cn=2Cn-Cn
= 1×2+3×2^2+5×2^3+...+(2n-3)×2^(n-1)+(2n-1)×2^n
-1×1-3×2-5×2^2-7×2^3-...-(2n-1)×2^(n-1)
=-1-2×[2+2^2+2^3+...+2^(n-1)]+(2n-1)×2^n
=-1-(2^2+2^3+2^4+,+2^n)+(2n-1)×2^n
=-1+[4-2^(n+1)]+(2n-1)×2^n
=-1+4-2×2^n+(2n-1)×2^n
=3+(2n-3)×2^n