数学卷5.31（22）：设数列{an}是等差数列,数列{bn}的前n项和Sn满足,Sn=1-bn,(n∈N*),a2-1

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数学卷5.31（22）：设数列{an}是等差数列,数列{bn}的前n项和Sn满足,Sn=1-bn,(n∈N*),a2-1=1/b1
a5=(1/b3)+1
（1）求数列{an}和{bn}的通项公式
（2）设Tn为数列{an▪bn}的前n项和,求Tn
求详解,要步骤.谢谢

s(n) = 1 - b(n),
b(1) = s(1) = 1 - b(1),
b(1)= 1/2.
s(n+1) = 1 - b(n+1),
b(n+1) = s(n+1) - s(n) = [1-b(n+1)] - [1-b(n)] = b(n)-b(n+1),
b(n+1) = b(n)/2,
{b(n)}是首项为b(1)=1/2,公比为1/2的等比数列.
b(n) = (1/2)(1/2)^(n-1) = 1/2^n.
a(n) = a + (n-1)d.
1/b(1)= 2 = a(2) - 1 = a + d - 1,a+d=3.
1 + 1/b(3) = 1 + 8 = 9 = a(5) = a+4d = a+d + 3d = 3+3d,d=2,a = 3-d = 1.
a(n) = 1 + 2(n-1) = 2n-1.
t(n) = a(1)b(1) + a(2)b(2) + a(3)b(3) + ...+ a(n-1)b(n-1) + a(n)b(n)
= (2*1-1)*(1/2) + (2*2-1)(1/2)^2 + (2*3-1)(1/2)^3 + ...+ [2(n-1)-1](1/2)^(n-1) + (2n-1)(1/2)^n,
2t(n) =(2*1-1) + (2*2-1)(1/2) + (2*3-1)(1/2)^2 + ...+ [2(n-1)-1](1/2)^(n-2) + (2n-1)(1/2)^(n-1),
t(n) = 2t(n) - t(n) = (2*1-1) + 2*(1/2) + 2*(1/2)^2+ ...+ 2*(1/2)^(n-1) - (2n-1)(1/2)^n
= -1 + 2[1 + 1/2 + ...+ (1/2)^(n-1)] - (2n-1)(1/2)^n
= -1 + 2[1 - 1/2^n]/(1-1/2) - (2n-1)(1/2)^n
= -1 + 4[1 - 1/2^n] - (2n-1)/2^n
= 3 - (2n+3)/2^n

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