Question

已知数列an满足an+1=an+2*3的n次方+1,a1=3,求数列an的通项公式

Answer 1

题目有歧义。如果正确的问题是a(n+1)=a(n)+2*3^n+1，则通项是a(n)=3^n+n-1.

Answer 2

a2=a1+2*3^1+1=10
a3=a2+2*3^2+1=29
...
所以不是等差或等比数列
a2=a1+2*3^1+1
a3=a2+2*3^2+1=a1+2*3^1+1+2*3^2+1=a1+2*(3^1+3^2)+1*2
a4=a3+2*3^3+1=a1+2*(3^1+3^2+3^3)+1*3
an+1=a1+2*(3^(n+1)-3)/2+1*n=3+(3^(n+1)-3)+n=3^(n+1)+n
an=3^n+(n-1)
n>=1
3^1+3^2+3^3....等比数列和
Sn=b1(1-q^n)/(1-q)=3*(1-3^n)/(1-3)=
(3^(n+1)-3)/2