Question

已知{an}是等差数列,其前n项和为sn, {bn}是等比数列,且a1=b1=2,a4+b4=27,

已知{an}是等差数列,其前n项和为sn, {bn}是等比数列,且a1=b1=2,a4+b4=27,s4-b4=10
(1) 求数列{an}与{bn}的通项公式

（2）记Tn=anb1+an-1b2+...+a1bn,n属于n正，求Tn。大神们求救啊？

Answer 1

a(n)=2+(n-1)d.
s(n)=2n+n(n-1)d/2.
b(n)=2q^(n-1).

10=s(4)-b(4)=8+6d-2q^3,
27=a(4)+b(4)=2+3d+2q^3,
37=10+9d, d=3.
a(n)=2+3(n-1)=3n-1.

10=8+6d-2q^3=26-2q^3,
q^3=8, q=2.
b(n)=2*2^(n-1)=2^n

（2）Tn=(3n-1)*2+(3n-4)*2^2+(3n-7)*2^3+......+8*2^(n-2)+5*2^(n-1)+2*2^n ①
2Tn=(3n-1)*2^2+(3n-4)*2^3+(3n-7)*2^4+......+8*2^(n-1)+5*2^n+2*2^(n+1)②
①-②，得Tn=-（3n-1)*2 + 3[2^2 + 2^3 + ... + 2^n] + 2^(n+2)
=2^(n+2) - 2(3n-1) + 12[1+2+...+2^(n-2)]
=2^(n+2)-2(3n-1)+12[2^(n-1)-1]
=2*2^(n+1)-6n+2 +3*2^(n+1)-12
=5*2^(n+1) - 6n - 10