数列{an}满足:an=3an-1+3n-1(n∈N,n≥2),其中a4=365,(1)求a1,a2,a3;(2)若{an+λ3n}为等差数列,求常数λ的值;(3)求{an}的前n项和Sn.-数学

题目简介

数列{an}满足:an=3an-1+3n-1(n∈N,n≥2),其中a4=365,(1)求a1,a2,a3;(2)若{an+λ3n}为等差数列,求常数λ的值;(3)求{an}的前n项和Sn.-数学

题目详情

数列{an}满足:an=3an-1+3n-1(n∈N,n≥2),其中a4=365,
(1)求a1,a2,a3; (2)若{
an
3n
}
为等差数列,求常数λ的值;(3)求{an}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(1)a1=5,a2=23,a3=95
(2)由{
an
3n
}
为等差数列可得:
an
3n
-
an-1
3n-1
为常数,
3n-(2λ+1)
3n
为常数,
所以2λ+1=0,
λ=-class="stub"1
2

(3)由2)可得an=(n+class="stub"1
2
)3n+1

Sn′=  class="stub"3
2
×3+class="stub"5
2
×32+…+ (n+class="stub"1
2
)3n

3Sn′=class="stub"3
2
×32+class="stub"5
2
×33+…(n-class="stub"1
2
3n+(n+class="stub"1
2
)
×3n+1
-2Sn′=class="stub"9
2
+32+33+…+3n-
(n+class="stub"1
2
3n+1

所以Sn=class="stub"n
2
(3n+1+1)

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