Question

数列an a1=-1 a2>a1 绝对值a(n+1)-an=2的n次方 若a(2n-1)单调递减

数列an a1=-1 a2>a1 绝对值a(n+1)-an=2的n次方 若a(2n-1)单调递减 a2n单调递增 数列an的通项公式为

Answer 1

|a(n+1) - a(n)| = 2^n,

2 = |a(2) - a(1)| = a(2) - a(1) = a(2) + 1, a(2) = 1.

a(2n-1)单调递减，a(2n-1) <= a(1) = -1 < 0.

a(2n)单调递增，a(2n)>= a(2) = 1 >0.

2^(2n) = |a(2n+1) - a(2n)| = a(2n) - a(2n+1),
2^(2n-1) = |a(2n) - a(2n-1)| = a(2n) - a(2n-1),

-2^(2n-1) = 2^(2n-1) - 2^(2n) = [a(2n) - a(2n-1)] - [a(2n) - a(2n+1)] = a(2n+1) - a(2n),
a(2n+1) = a(2n) - 2^(2n-1) = a(2n-1) - 2*2^(2n-2) = a(2n-1) - 2*4^(n-1),
a(2n+1)/4^n = (1/4)a(2n-1)/4^(n-1) - 1/2,
a(2n+1)/4^n + 2/3 = (1/4)[a(2n-1)/4^(n-1)] + 1/6 = (1/4)[ a(2n-1)/4^(n-1) + 2/3 ]
{a(2n-1)/4^(n-1) + 2/3}是首项为a(1) + 2/3 = -1/3, 公比为(1/4)的等比数列。
a(2n-1)/4^(n-1) + 2/3 = (-1/3)(1/4)^(n-1),

a(2n-1) + (2/3)4^(n-1) = -1/3,
a(2n-1) = -1/3 -(2/3)4^(n-1) = -1/3 - (2/3)2^(2n-2) = -1/3 - (1/3)2^(2n-1)

a(2n) = 2^(2n-1) + a(2n-1) = 2*2^(2n-2) + a(2n-1) = 2*4^(n-1) - 1/3 - (2/3)4^(n-1)
= (4/3)4^(n-1) - 1/3
= -1/3 + (1/3)4^n
= -1/3 + (1/3)2^(2n).

a(2n-1) = -1/3 - (1/3)2^(2n-1) = -1/3 + (1/3)(-2)^(2n-1)
a(2n) = -1/3 + (1/3)2^(2n) = -1/3 + (1/3)(-2)^(2n)

{a(n)}的通项公式为，
a(n) = -1/3 + (1/3)(-2)^n,