解
(1) a(n+1)=-an+3^n
兩邊同時(shí)除 3^n 得
a(n+1) / 3^n= - an / 3^n+1
這里注意變形

a(n+1) / 3^n= (-1/3) *3 * an / 3^n+1
a(n+1) / 3^n= (-1/3) * an / 3^(n-1)+1
設(shè)數(shù)列 {bn}令 bn =an / 3^(n-1)得
b(n+1)= (-1/3)bn +1其中 b1 =a1 / 3^(n-1) =0
兩邊同時(shí) 減3/4(這里用了構(gòu)造法中 X =p / q-1 ,即
1 / [(-1/3)-1] ) 得
b(n+1) -3/4= (-1/3)bn +1- 3/4

b(n+1) -3/4= (-1/3)bn + 1/4

b(n+1) -3/4= (-1/3)(bn- 3/4)
所以{bn- 3/4} 是首項(xiàng)為 -3/4公比為-1/3 的等比數(shù)列
即bn - 3/4=(-3/4)* (-1/3)^(n-1)

bn=(-3/4)* (-1/3)^(n-1)+ 3/4
因?yàn)閎n = an / 3^(n-1)

所以an=[(-3/4)* (-1/3)^(n-1)+ 3/4]* 3^(n-1)
最后化簡(jiǎn)得an = (3/4)[3^(n-1) + (-1)^n]
(2) 沒(méi)有最大值只有最小值
最小值由通項(xiàng)可以知道

當(dāng) n = 1 時(shí)an最小a1=0