已知数列an相邻两项an,an+1是方程X^2-（2^n）*X+bn=0的两实根,a1=1.求证数列an-（1/3）*（2^n）是等比数列,并求an的通项公式

an,a(n+1)是方程X^2-（2^n）*X+bn=0的两实根,所以,an+a(n+1)=2^n.
所以,a(n+1)=2^n - an = 2^n - [ 2^(n-1) - a(n-1) ] = 2^n - 2^(n-1) + a(n-1) = 2^(n-1) + a(n-1).
a1=1,a2=2^1-a1=1,以下分奇偶数讨论.
1.当n=2k,为偶数时,
a(2k)=2^(2k-2) + a(2k-2) = 2^(2k-2) + 2^(2k-4) + a(2k-4) = .
= 2^(2k-2) + 2^(2k-4) +.+2^(2) + a2 = 2^(2k-2) + 2^(2k-4) +.+2^2 + 2^0
= Σ_(i=1) ^(k) 2^(2i-2) = Σ_(i=1) ^(k) 4^(i-1) = 1*(1-4^k) / (1-4) = (4^k - 1) /3
= [2^(2k) - 1] /3 = (2^n - 1) /3
2.当n=2k+1,为奇数时,
a(2k+1)=2^(2k-1) + a(2k-1) = 2^(2k-1) + 2^(2k-3) + a(2k-3) = .
= 2^(2k-1) + 2^(2k-3) +.+2^(1) + a1 = 2^(2k-1) + 2^(2k-3) +.+2^1 + 1
= 1+Σ_(i=1) ^(k) 2^(2i-1) =1 + 2*(1-4^k) / (1-4) =1+ 2 (4^k - 1) /3
= [2^(2k+1) + 1] /3 = (2^n + 1) /3
所以,an -(1/3)*(2^n) = 1/3 当n为奇数时,
= -1/3 当n为偶数时,
是公比为-1 的等比数列.
而an的通项公式为 an = (2^n + 1) /3 当n为奇数时,
= (2^n - 1) /3 当n为偶数时.