Question

an是等差数列,bn是各项都为正数的等比数列,a1=b1,a3+b5=21,a5+b3=13,求an乘bn的前n项和sn

Answer 1

设an=1+(n-1)d,bn=q^(n-1)

则1+2d+q^4=21

1+4d+q^2=13

解得d=2

q=2

所以 an=2n-1,bn=2^(n-1)

令Sn为数列an/bn的前项和,则

Sn=1/1+3/2+5/4+……+(2n-1)/2^(n-1) ①

两边同乘以2,得

2Sn=2/1+3/1+5/2+……+(2n-1)/2^(n-2) ②

②-①【错位相减】,得

Sn=2/1+2/1+2/2+……+2/2^(n-2)-(2n-1)/2^(n-1)

=2+2*［1/1+1/2+1/4+……+1/2^(n-2)］-(2n-1)/2^(n-1)

=2+2*［1-(1/2)^(n-1)］/(1-1/2)- (2n-1)/2^(n-1)

=2+4*［1-(1/2)^(n-1)］-(2n-1)/2^(n-1)

=2+4-4/2^(n-1)-(2n-1)/2^(n-1)

=6-(4+2n-1)/2^(n-1)

=6-(2n+3)/2^(n-1)