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=19,a5+b3=9.

a1=b1=1
a3+b5=1+2d+q^4=19
2d+q^4=18
2d=18-q^4
d=9-q^4/2
a5+b3=1+4d+q^2=9
4d+q^2=8
4d=8-q^2
d=2-q^2/4
9-q^4/2=2-q^2/4
7-q^4/2+q^2/4=0
2q^4-q^2-28=0
(q^2-4)(2q^2+7)=0(2q^2+7>0)
q^2=4{bn}是各项都为正数
q>0
q=2
d=2-q^2/4=2-2^2/4=1
an=a1+(n-1)d
=1+(n-1)
=n
bn=b1q^n-1
=2^n-1
a1=1
a2=2
a3=3
...
an=n
b1=1
b2=2
b3=4
...
bn=2^n-1
S=1*2^0+2*2^1+3*^2+4*2^3+.+n*2^n-1
2S=1*2^1+2*2^2+3*2^3+.+n*2^n
2S-s=-2^0-2^1-2^2-2^3-.2^n-1+n*2^n
=-(2^0+2^1+2^2+2^3+.+2^n-1)+n*2^n
=-[(1-2^n)/(1-2)]+n*2^n
=1-2^n+n*2^n
=1+(n-1)*2^n

是a1+b1=1，还是a1=b1=1？搂主写的是前者，一楼做的是后者。