已知数列{an}满足：a1=1

Question

如题所述

Answer 1

0 = [3+(-1)^n]a(n+2) - 2a(n) + 2[(-1)^n - 1],

0 = [3+(-1)^(2n-1)][a(2n-1+2) - 2a(2n-1) + 2[(-1)^(2n-1) - 1]
= (3-1)a(2n+1) - 2a(2n-1) + 2[-1-1]
= 2a(2n+1) - 2a(2n-1) - 4,
a(2n+1) = a(2n-1) + 2,
{a(2n-1)}是首项为a(1)=1,公差为2的等差数列。
a(2n-1) = 1 + 2(n-1) = 2n-1.

0 = [3+(-1)^(2n)]a(2n+2) - 2a(2n) + 2[(-1)^(2n)-1]
= (3+1)a(2n+2) - 2a(2n) + 2[1-1]
= 4a(2n+2) - 2a(2n).
a(2n+2) = a(2n)/2.
{a(2n)}是首项为a(2)=1/2,公比为1/2的等比数列。
a(2n) = (1/2)*(1/2)^(n-1) = 1/2^n = [1/2^(1/2)]^(2n).

a(n)的通项公式为，
n为奇数时，a(n) = n.
n为偶数时，a(n) = [1/2^(1/2)]^n.
a(3) = 3, a(5) = 5.
a(4) = 1/2^2 = 1/4, a(6) = 1/2^3 = 1/8.

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

s(n) = 2s(n)-s(n) = (2*1-1) + 2/2 + 2/2^2 + ... + 2/2^(n-1) - (2n-1)/2^n
= -1 + 2[1+1/2+1/2^2 + ... +1/2^(n-1)] - (2n-1)/2^n
= 2[1 - 1/2^n]/(1-1/2) - 1 - (2n-1)/2^n
= 4[1-1/2^n] - 1 - (2n-1)/2^n
= 3 - (2n+3)/2^n