Question

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

求，【1】{an}{bn}的通项公式【2】{anbn}的前n项和Sn

Answer 1

(1).a3+b5=a1+2d+b1q^4=21
a5+b3=a1+4d+b1q^2=13
d=2
q=2
an=1+2(n-1)=2n-1
bn=2^(n-1)
（2）
cn=anbn=(2n-1)*2^(n-1)
c1=1*2^0
c2=3*2^1
c3=5*2^2
....cn=(2n-1)*2^(n-1)
Sn =c1+c2+c3+......cn
=1*2^0+3*2^1+5*2^2+.....+(2n-1)*2^(n-1)
2Sn= 1*2^1+3*2^2+5*2^3+......+(2n-1)*2^n
2Sn-Sn=Sn=-1-2*2^1-2*2^2-2*2^3-.......-2*2^(n-1)+(2n-1)*2^n
Sn=-1-2(2^1+2^2+2^3+....+2^(n-1))+(2n-1)*2^n
=-1-2(2^n-2)+(2n-1)*2^n
=3-2^(n+1)+(2n-1)*2^n
=3+(2n-3)*2^n
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