数列求前n项和设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=132)求数列{an/bn}的前n项和Sn

Question

数列求前n项和
设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13
2)求数列{an/bn}的前n项和Sn

Answer 1

1.设an=a1+(n-1)d=1+(n-1)d
bn=b1*q^(n-1)=q^(n-1)
a3+b5=1+2d+q^4=21 ①
a5+b3=1+4d+q^2=13 ②
联立①②得q^2=4
因为{bn}各项为正数
所以q=2 则d=2
an=2n-1
bn=2^(n-1)
2.an/bn=(2n-1)/[2^(n-1)]=(4n-2)/2^n=(4n/2^n)-(2/2^n)
设cn=4n/2^n dn=2/2^n前N项和为Cn,Dn
Cn=4/2+4*2/2^2+4*3/2^3...+4n/2^n ①
1/2Cn=4/2^2+4*2+2^3+...+4(n-1)/2^n+4n/2^(n+1) ②
①-②得1/2Cn=4/2+4/2^2+4/2^3+...+4/2^n-4n/2^(n+1)
=2{[1-(1/2)^n]/(1-1/2)}-4n/2^(n+1)=4[1-(1/2)^n]-4n/2^(n+1)
Dn=2/2+2/2^2+2/2^3+...+2/2^n[1-(1/2)^n]/(1-1/2)=2[1-(1/2)^n]
Sn=Cn-Dn=2[1-(1/2)^n]-4n/2^(n+1)