Question

1、在数列{an}中，a1=1.a(n+1)=3an+2n+1.求an. 2、在数列{an}中，a1=-1,a(n+1)=(3an-4)/[(an)-1].求an.

Answer 1

a(n+1)=3a(n)+2n+1=3a(n)+3n-(n+1)+2=3a(n)+3n-(n+1)+3-1,
a(n+1)+(n+1)+1=3[a(n)+n+1],
{a(n)+n+1}是首项为a(1)+1+1=3,公比为3的等比数列.
a(n)+n+1=3*3^(n-1)=3^n.
a(n)=3^n - n - 1.

2,
a(n+1)=[3a(n)-4]/[a(n)-1],
a(n+1)-2=[3a(n)-4]/[a(n)-1]-2=[3a(n)-4-2a(n)+2]/[a(n)-1]=[a(n)-2]/[a(n)-1],
若a(n+1)=2,则a(n)=2, ..., a(1)=2,与a(1)=-1矛盾,因此,a(n)不为2.
1/[a(n+1)-2] = [a(n)-1]/[a(n)-2]=[a(n)-2+1]/[a(n)-2] = 1/[a(n)-2] + 1
{1/[a(n)-2]}是首项为1/[a(1)-2]=1/[-1-2]=-1/3,公差为1的等差数列.
1/[a(n)-2] = -1/3 + (n-1) = (3n-4)/3,
a(n)-2=3/(3n-4),
a(n)=2+3/(3n-4)=(6n-8+3)/(3n-4)=(6n-5)/(3n-4)

Answer 2

1
∵a(n+1)=3an+2n+1.
a(n+1)+n+2=3(an+n+1)
∴[a(n+1)+n+2]/(an+n+1)=3
∴{an+n+1}为等比数列，公比为3
∴an+n+1=(a1+2)3^(n-1)=3^n
∴an=3^n-1-n
2
∵a(n+1)=(3an-4)/(an-1)
∴a(n+1)-2=(3an-4)/(an-1)-2
a(n+1)-2=(an-2)/(an-1)
两边取倒数
1/[a(n+1)-2]=(an-1)/(an-2)=1+1/(an-2)
∴1/[a(n+1)-2]-1/(an-2)=1
∴{1/(an-2)}为等差数列，公差为1
1/(an-2)=1/(a1-2) +n-1=n-4/3
∴an-2=1/(n-4/3)
∴an=2+3/(3n-4)

Answer 3

1.an=3^n-n-1