Question

已知an是公差d不为0的等差数列,a1=1且a1,a3,a9成等比数列(1)求an通项公(2)bn=(2an

已知an是公差d不为0的等差数列，a1=1且a1,a3,a9成等比数列(1)求an通项公(2)bn=(2an+1)2^an求bn前n项和(3)设cn=(1/2)^n(2^an+1)【2^（an+1）+1】，求Tn=1/c1+1/c2+...+1/cn,并求Tn的最小值

Answer 1

a(n)=1+(n-1)d, d不为0。
[a(3)]^2 = [1+2d]^2 = a(1)a(9) = 1+8d = 1 + 4d + 4d^2 , 0 = 4d^2 - 4d = 4d(d-1), d=1.
a(n) = 1 + (n-1) = n.

b(n) = [2a(n)+1)2^[a(n)] = (2n+1)2^n,
s(n) = b(1)+b(2)+b(3)+...+b(n-1)+b(n)
= (2*1+1)2 + (2*2+1)2^2 + (2*3+1)2^3 + ... + [2(n-1)+1]2^(n-1) + [2n+1]2^n,
2s(n) = (2*1+1)2^2 + (2*2+1)2^3 + ... + [2(n-1)+1]2^n + [2n+1]2^(n+1),
s(n) = 2s(n)-s(n) = -(2*1+1)2 - 2*2^2 - 2*2^3 - ... - 2*2^n + [2n+1]2^(n+1)
= (2n+1)2^(n+1) - 2[1+2+2^2+...+2^n]
= (2n+1)2^(n+1) - 2[2^(n+1)-1]/(2-1)
= (2n+1)2^(n+1) - 2*2^(n+1) + 2
= (2n-1)2^(n+1) + 2,

c(n) = (1/2)^n{2^[a(n)] + 1}{2^[a(n)+1] + 1} = (1/2)^n[2^n + 1}{2^(n+1)+1},
1/c(n) = 2^n/{[2^n + 1][2^(n+1) + 1]} = 1/[2^n + 1] - 1/[2^(n+1) + 1],
t(n) = 1/c(1) + 1/c(2) + 1/c(3) + ... + 1/c(n-1) + 1/c(n)
= 1/[2+1]-1/[2^2+1] + 1/[2^2+1]-1/[2^3+1] + 1/[2^3+1]-1/[2^4+1] + ... + 1/[2^(n-1)+1]-1/[2^n+1] + 1/[2^n + 1]-1/[2^(n+1)+1]
= 1/[2+1] - 1/[2^(n+1)+1]
= 1/3 - 1/[2^(n+1)+1],
t(n)单调递增，t(n)>=t(1) = 1/3 - 1/[2^2 + 1] = 1/3 - 1/5 = 2/15,
t(n)的最小值为2/15

Answer 2

an＝n
bn的前n项和为2∧（n＋1）