题目

已知数列{an}满足a1=4,an+1=an+p.3^n+1(n属于N+,P为常数）,a1,a2+6,a3成等差数列.
（1）求p的值及数列{an}的通项公式.
(2)设数列{bn}满足bn=n^2/(an-n),证明：bn<=4/9.

提问时间：2020-12-07

答案

经化简得a1 a2 a3 分别为a1=4 a2=a1+3p+1=5+3p  a3=a1+12p+2=6+12pa1,a2+6,a3成等差数列.的2a2+12=a1+a3 即22+6p=10+12p 解得p=2a(n+1)=an+2*3^n+1a2-a1=2*3^1+1a3-a2=2*3^2+1...an-a(n-1)=2*3^(n-1)+1将这些式...