数列{an}满足递推公式an=3an-1+3n-1(n≥2),又a1=5,则使得{an+λ3n}为等差数列的实数λ=( ) A.2 B.5 C.-12 D.12
问题描述:
数列{an}满足递推公式an=3an-1+3n-1(n≥2),又a1=5,则使得{
}为等差数列的实数λ=( )
an+λ 3n
A. 2
B. 5
C. -
1 2
D.
1 2
答
设bn=an+λ3n,根据题意得bn为等差数列即2bn=bn-1+bn+1,而数列{an}满足递推式an=3an-1+3n-1(n≥2),可取n=2,3,4得到3a1+32−1+λ32+3a3+34−1+λ34=23a2+33−1+λ33,而a2=3a1+32-1,a3=3a2+33-1=3(3a1+32-1)...