a_1 = 1
用展开法
a_n =3(a_(n-1) +1)=3^2 (a_(n-2)+1) +3 = 3^3 (a_(n-3)+1)+3^2+3
=......
=3^(n-1) (a_(n-(n-1))+1) +3^(n-2)+3^(n-3) +......+3^2 +3
=2x3^(n-1) +3^(n-2)+3^(n-3) + ... + 3^2 + 3
=2x3^(n-1) +3 x ( 3^(n-3) + ... + 3^2 + 3 +1)
=2x3^(n-1) + 3 x ( (1-3^(n-2)) / (1-3) )
=2x3^(n-1) + 3 x (3^(n-2) -1)/2 )
=2x3^(n-1) + (3^(n-1) -3)/2
=(5/2) 3^(n-1) -3/2
设 an+₁ +A=3(an+A)
则 A=½
设数列 bn=an+½
则 数列{bn} 为初项 b₁=3/2,公比为3的等比数列
bn=b₁•3ⁿˉ¹=3ⁿ/2
即 an=bn-½=½•(3ⁿ-1)
两边同加2,然后构造数列
提抄错了