已知等差数列{an}公差d不等于零,其前n项和为Sn,且7a5=2a8+9,a1,a3,a7成等比数例（1）求an通项公

a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2,
[a(3)]^2 = [a+2d]^2 = a(1)a(7)=a(a+6d) = a^2 + 6ad = a^2 + 4ad + 4d^2,
0 = 4d^2 - 2ad = 2d(2d-a), a = 2d.
a(n)=2d+(n-1)d=(n+1)d,
7a(5)=7*6d=2a(8)+9=2*9d + 9 = 18d + 9 = 42d,
9 = 24d, d = 9/24 = 3/8.
a(n)=3(n+1)/8.
a(n)=(3/16)[(n+1)(n+2)-n(n+1)],
s(n)=(3/16)[(n+1)(n+2)-1*2]=(3/16)[(n+1)(n+2)-2]
b(2n-1)=a(2n-1)=3(2n-1+1)/8=3n/4.
b(2n)=2^(2n)s(2n)/(2n)=2^(2n)(3/16)[(2n+1)(2n+2)-2]/(2n)=(3/16)4^n[(2n+1)(n+1)-1]/n
=(3/16)4^n[2n(n+1)+n]/n
=(3/16)4^n[2(n+1)+1]
=(3/16)4^n(2n+3),
t(2n)=b(1)+b(3)+...+b(2n-1)+b(2)+b(4)+...+b(2n)=A(n)+B(n),
A(n)=(3/4)(1+2+...+n)=(3/4)n(n+1)/2=3n(n+1)/8.
B(n)=(3/16)[4(2*1+3)+4^2(2*2+3)+...+4^(n-1)(2n+1)+4^n(2n+3)],
4B(n)=(3/16)[4^2(2*1+3)+4^3(2*2+3)+...+4^n(2n+1)+4^(n+1)(2n+3)],
3B(n)=4B(n)-B(n)=(3/16)[4^(n+1)(2n+3)-4(2*1+3)-4^2*2-...-4^n*2]
=(3/16)[4^(n+1)(2n+3)] - (3/8)[4+4^2+...+4^n]- (3/16)(4/3)
=(3/16)4^(n+1)(2n+3) - (3/2)[1+4+...+4^(n-1)] - 3/12
=3(2n+3)4^(n-1) - (3/2)[4^n - 1]/(4-1) - 3/12
=3(2n+3)4^(n-1) - (3/6)[4^n - 1] - 3/12,
B(n)=(2n+3)4^(n-1) - [4^n - 1]/6 - 1/12
=(2n+3)4^(n-1) - (2/3)4^(n-1) + 1/12
=[(6n+7)/3]4^(n-1) + 1/12
t(2n)=A(n)+B(n)=3n(n+1)/8 + [(6n+7)/3]4^(n-1) + 1/12,
t(2n-1) = t(2n)-b(2n)=3n(n+1)/8 + [(6n+7)/3]4^(n-1) + 1/12 - (3/16)4^n(2n+3)
= 3n(n+1)/8 + [(6n+7)/3]4^(n-1) + 1/12 - [(6n+9)/4]4^(n-1)
= 3n(n+1)/8 + [(24n+28-18n-27)/12]4^(n-1) + 1/12
= 3n(n+1)/8 + [(6n+1)/12]4^(n-1) + 1/12