an=-a（n－1）+2n，a1等于1，求an通项公式 为什么有时候数列的待定系数法不能用？
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a1=1
an=-a(n-1)+2n
let
an + k1.n +k2 = - [ a(n-1) + k1.(n-1) +k2 ]
coef. of n
-2k1 = 2
k1 =-1
coef. of constant
k1-2k2 = 0
-1-2k2=0
k2 =-1/2
ie
an=-a(n-1)+2n
an - n -1/2 = - [ a(n-1) -(n-1) -1/2 ]
=>{an - n -1/2} 是等比数列, q=-1
an - n -1/2 =(-1)^(n-1) . (a1 - 1 -1/2)
=(1/2).(-1)^n
an = n +1/2 +(1/2).(-1)^n
an=-a(n-1)+2n
let
an + k1.n +k2 = - [ a(n-1) + k1.(n-1) +k2 ]
coef. of n
-2k1 = 2
k1 =-1
coef. of constant
k1-2k2 = 0
-1-2k2=0
k2 =-1/2
ie
an=-a(n-1)+2n
an - n -1/2 = - [ a(n-1) -(n-1) -1/2 ]
=>{an - n -1/2} 是等比数列, q=-1
an - n -1/2 =(-1)^(n-1) . (a1 - 1 -1/2)
=(1/2).(-1)^n
an = n +1/2 +(1/2).(-1)^n
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