数列an满足递推式an=3an-1+3^n-1,n大于等于2,其中a1=5,则使得{an+入、3……n}为等差数列的实数入=-------------
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/07 18:06:10
数列an满足递推式an=3an-1+3^n-1,n大于等于2,其中a1=5,则使得{an+入、3……n}为等差数列的实数入=-------------
数列an满足递推式an=3an-1+3^n-1,n大于等于2,其中a1=5,则使得{an+入、3……n}为等差数列的实数入=-------------
数列an满足递推式an=3an-1+3^n-1,n大于等于2,其中a1=5,则使得{an+入、3……n}为等差数列的实数入=-------------
真佩服你,题目也错了应该是{an+入/3^n}是等差数列
an=3a(n-1)+3^n (两边同除以3^n)
an/3^n=a(n-1)/3^(n-1)+1
an/3^n-a(n-1)/3^(n-1)=1
所以{an/3^n}是以a1/3=5/3为首项d=1为公差的等差数列
入是脚标呢还是分子项,结果不一样
已知数列{an}满足a1=1 an+1=an/(3an+1) 则球an
数列{an}满足a1=1,且an=an-1+3n-2,求an
数列[An]满足An+1-An+3=0,且A1=-5.求An.
已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列
已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an
已知数列{an}满足a1=1/2,an+1=3an+1,求数列{an}通项公式
数列an满足a1=1/2 an+1=an/(2an+3) 猜想数列通项公式
已知数列[an]满足An+1=1+an /3-an ,且a1=1/3,求证数列[1/(an -1)]是等差数列,并求an
数列an满足a1=1,a2=1/3,并且an(an-1+an+1)=2an+1an-1.则数列第2012项为?
数列{an}满足a1=a,an+1=1+1/an.若3/2
数列an满足a1=1/3,Sn=n(2n-1)an,求an
数列{an}满足Sn+Sn+1=5/3an+1,a1=4求an
已知数列{an}满足3an+1+an=4,a1=9,求通项公式.
数列[An]满足a1=2,a(n+1)=3an-2 求an
已知数列an满足a1=0,an+1=an-根号3/根号3an+1,则a2012=
已知数列{an}满足a1=4/3,且an+1=4(n+1)an/3an+n
已知数列an满足a1=1,a2=3,an+1.an-1=an,求a2013
已知数列an满足a1=1,a(n+1)=an/(3an+2),则an=?