✔ 最佳答案
Let P(n) be the statement a_n=(1/3)[2^n-(-1)^n]
when n=1, a_1=1=(1/3)[2-(-1)]. P(1) is true
when n=2, a_1=1=(1/3)[2^2-(-1)^2]. P(2) is true
Assume that when n=k, P(k) is true, when n=k+1
a_k+1=a_k+a_k-1
=(1/3)[2^k-(-1)^k]+(2/3)[2^(k-1)-(-1)^(k-1)]
=(1/3)[2^k-(-1)^k+2^k-(-1)^2(k-1)]
=(1/3)[2^(k+1)+(-1)^(k+1)]
So P(k+1) is true. By M.I. for all positive integer n, a_n=(1/3)[2^n-(-1)^n]
2010-01-16 22:26:19 補充：
This is just the second order difference equation. Consider the chacteristic equation
λ^2=λ+2=>λ^2-λ-2=0=>(λ-2)(λ+1)=0
=>λ=2 or -1
So a_n=A2^n+B(-1)^n. Using a_1=a_2=1
2A-B=1,4A+B=1=>A=1/3,B=-1/3
So a_n=(1/3)[2^n-(-1)^n]. OK ?
