1)设{an}的公差为d,{bn}的公比为q
a2+b2=10
a1+d+b1q=10
1+d+4q=10
d+4q=9 ①
a26-b3=10
a1+25d-b1q^2=10
1+25d-4q^2=10
25d-4q^2=9 ②
解① ②方程组,得q=2,d=1
an=1+(n-1)*1=n
bn=4*2^(n-1)=2^(n+1)
2)cn=anbn
=n*2^(n+1)
Sn=1*2^2+2*2^3+3*2^4+……+n*2^(n+1) ③
2Sn=1*2^3+2*2^4+3*2^5+……+n*2^(n+2) ④
③- ④ 【错位相减】,得
-Sn=2^2+2^3+2^4+……+2^(n+1) -n*2^(n+2)
=2^2(2^n-1)/(2-1)-n*2^(n+2)
=2^(n+2)-2^2-n*2^(n+2)
=(1-n)2^(n+2)-4
Sn=(n-1)2^(n+2)+4