Question

设an是各项都为正数的等比数列，bn是等差数列，且a1=b1=1.a3+b5=13.a5+b3=2

设an是各项都为正数的等比数列，bn是等差数列，且a1=b1=1.a3+b5=13.a5+b3=21.设an的前n项和为sn。求数列sn×bn的前n项和Tn

Answer 1

(1)设公差为d，公比为q，显然q>0
则2d+q^4=20 (1) 4d+q^2=12 (2)
(1)*2-(2) (2q^2+7)(q^2-4)=0
∵q>0
∴q=2 代入得d=2
an=1+2(n-1)=2n-1
bn=2^(n-1)
(2)Sn=1+3/2+5/2^2+....+(2n-1)/2^(n-1) (3)
2Sn=2+3+5/2+.....+(2n-1)/2^(n-2) (4)
(4)-(3) Sn=2+2+2/2+...+2/2^(n-2)-(2n-1)/2^(n-1)
=4+[1+1/2+...+1/2^(n-1)]-(2n-1)/2^(n-1)
=4+[1-1/2^(n-1)]/(1-1/2)-(2n-1)/2^(n-1)
=4+2-2/2^(n-1)-(2n-1)/2^(n-1)
=6-(2n+1)/2^(n-1)