数列[An]满足An+1-An+3=0,且A1=-5.求An.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/24 00:23:01
x){6uӎюyv/|mc.05ycMyz6IE$/!ÆK^.l';zP&TY-zO[;bg ?ӞiOv(u5~dR<~I/M{p'H5OufB@8:jhM55ttn(v̫|0b&X<;~
数列[An]满足An+1-An+3=0,且A1=-5.求An.
数列[An]满足An+1-An+3=0,且A1=-5.求An.
数列[An]满足An+1-An+3=0,且A1=-5.求An.
此题是不是A(n+1)-An+3=0的.先当这样,此式可化为A(n+1)=An-3所以此数列是首项为A1,公差为-3的数列,所以An=A1+d(n-1)=-5-3(n-1)=-3n-2,即数列{An}的通项为-3n-2.
数列[An]满足An+1-An+3=0,且A1=-5.求An.
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=p/(pn+1-p)(0
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=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
数列{an)满足an=4a(n-1)+3,a1=0,求数列{an}的通项公式
已知数列an满足a1=1,1/an+1=根号1/an^2+2,an>0,求an
数列{an}满足:a1=1,an>0,an+1^2-an^2=1,那么an
已知数列{an}满足a1=1,an+1 -an+2an+1•an=0求通项
数列{an}满足:a1=1,an>0,an+1^2-an^2=1,那么an
已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an
数列an满足,a1=8且8an+1an-16an+1+2an+5=0
数列an满足an+1=根号(an^2+1)+an,a1=a>0,求an通项公式
已知数列{an}满足,a1=2,a(n+1)=3根号an,求通项an数列{an}满足:an>0,且根号下Sn=an+1/4,求通项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
数列{an}满足an+1=2an 0≤an≤1/2 an+1=2an-1 1/2≤an