Question

用待定系数法求数列通项公式！！！要详细步骤

Answer 1

简单分析一下，详情如图所示

Answer 2

(1)
a1=1
2a(n+1)= 3an +3
a(n+1) = (3/2)an + 3/2
a(n+1) + 3 = (3/2)(an + 3 )
=> {an + 3} 是等比数列, q=3/2
an + 3 = (3/2)^(n-1) .(a1 + 3)
an = -3 + 4.(3/2)^(n-1)

(2)
a1=1
a(n+1) = 3an +2
a(n+1) + 1 = 3(an + 1)
=>{an + 1} 是等比数列, q=3
an + 1= 3^(n-1) .( a1 +1)
an = -1 + 2. 3^(n-1)

(3)
a1=1
a(n+1) = 2an +1
a(n+1) + 1 = 2( an + 1)
=>{an + 1} 是等比数列, q=2
an +1 =2^(n-1) . (a1+1)
an = -1+ 2^n