An>0,A1=2,当n>=2,
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/03 06:44:28
An>0,A1=2,当n>=2,
An+A(n-1)=n/(An-A(n-1))+2,求An通项
An+A(n-1)=n/(An-A(n-1))+2,求An通项
[An+A(n-1)][An-A(n-1)]=n+2[An-A(n-1)]
A²n-A²(n-1) =n+2[An-A(n-1)]
所以
A²2-A²1 =2+2(A2-A1)
A²3-A²2 =3+2(A3-A2)
...
A²n-A²(n-1) =n+2[An-A(n-1)]
左右2边全部相加得
A²n-A²1=(n-1)(n+2)/2 +2(An-A1)
A²n-4=(n-1)(n+2)/2+2An-4
A²n-2An+1=(n-1)(n+2)/2 +1 = (n²+n-2+2)/2
(An-1)²=n(n+1)/2
An= 1+根号[n(n+1)/2]
A²n-A²(n-1) =n+2[An-A(n-1)]
所以
A²2-A²1 =2+2(A2-A1)
A²3-A²2 =3+2(A3-A2)
...
A²n-A²(n-1) =n+2[An-A(n-1)]
左右2边全部相加得
A²n-A²1=(n-1)(n+2)/2 +2(An-A1)
A²n-4=(n-1)(n+2)/2+2An-4
A²n-2An+1=(n-1)(n+2)/2 +1 = (n²+n-2+2)/2
(An-1)²=n(n+1)/2
An= 1+根号[n(n+1)/2]
An>0,A1=2,当n>=2,
数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公
An>0,A1=2,当n>=2,An+A(n-1)=n/(An-A(n-1))+2,求An通项
已知数列an中满足a1=1且当n.=2时,2an*a*(n-1)+an-a(n-1)=0,求通项公式an
已知数列{an}中,a1=1,当n>等于2时,an+2SnSn-1=0
数列an中,a1=1/4 ,当n>=2时,有(3n^2-2n-1)an=a1+a2+a3+.+a(n-1)
数列{an}中,a1=1,当n>=2时,Sn=n2an (n的平方*an),求通项an.a1=1不是=1/2.
1:在数列{an}中,a1=1,当n>=2时,其前n项和sn满足an+2sn*s(n-1)=0
已知an=(-1)的n次方,当n=1时,a1=0; 当n=2时,a2=2,当n=3时,a3=0,...则a1+a2+a3
数列{an}中,a1=1,当n>1时,2Sn^2=2anSn-an,求通项an
数列{an}中,a1=1,当n>1时2Sn²=2anSn-an,求通项an
已知数列{an}满足an+1+an=4n-3 当a1=2时,求Sn