数列an满足a(n+1)=|an-1| ,a1=5/4,求an

a1 = 5/4
a2 = |5/4 - 1| = 1/4
a3 = |1/4 - 1| = 3/4
a4 = |3/4 - 1| = 1/4
a5 = |1/4 - 1| = 3/4
...
a1 = 5/4,
a(2n) = 1/4,
a(2n+1) = 3/4, n = 1,2,...

利用归纳法证明，
1）n = 1时，a(2) = 1/4, a(3) = 3/4.结论成立。
2）假设，对n = k时，有 a(2k) = 1/4, a(2k+1) = 3/4.
则 a[2(k+1)] = a(2k+2) = a(2k+1+1) = |a(2k+1) - 1|
= |3/4 - 1| = 1/4,
a[2(k+1)+1] = |a[2(k+1)] - 1| = |1/4 - 1| = 3/4.
因此，
若当n = k时，有 a(2k) = 1/4, a(2k+1) = 3/4.
则，当n = k+1时，有 a[2（k+1)] = 1/4, a[2(k+1)+1] = 3/4.

所以，
由归纳法知，
a(1) = 5/4,
a(2n) = 1/4,
a(2n+1) = 3/4, n = 1,2,...

An=1/4(n为偶数)
An=3/4(n奇数）

不是很难