1.a1=3 且 a(n+1)=an+5ana(n+1)
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1.a1=3 且 a(n+1)=an+5ana(n+1)
2.a1=1 且 a(n+1)=an+2n+1
3.a1=1 且 a(n+1)=an+1/4n^2-1
4.a1=1 an=n+1/n*a(n+1)
求各题的an
2.a1=1 且 a(n+1)=an+2n+1
3.a1=1 且 a(n+1)=an+1/4n^2-1
4.a1=1 an=n+1/n*a(n+1)
求各题的an
1.a1=3 且 a(n+1)=an+5ana(n+1)
a(n+1)=an+5ana(n+1)
1/an=1/a(n+1)+5
1/a(n+1)-1/an=-5
1/an=1/a1-5(n-1)=1/3-5(n-1)=(16-15n)/3
an=3/(16-15n);
2.a1=1 且 a(n+1)=an+2n+1
a(n+1)-an=2n+1
an-a(n-1)=2(n-1)+1
a(n-1)-a(n-2)=2(n-2)+1
……
a3-a2=2*2+1
a2-a1=2*1+1
两边相加:
an-a1=[2(n-1)+1]+[2(n-2)+1]+[2(n-3)+1]+……+[2*2+1]+[2*1+1]
=2[(n-1)+(n-2)+(n-3)+……+2+1]+(n-1)
=n(n-1)+(n-1)
=(n+1)(n-1)
an=a1+(n+1)(n-1)=n^2;
3.a1=1 且 a(n+1)=[(an+1)/4]n^2-1
a(n+1)=[(an+1)/4]n^2-1
a(n+1)+1=[(an+1)/4]n^2
[a(n+1)+1]/(an+1)=(1/4)n^2
[an+1]/[a(n-1)+1]=(1/4)(n-1)^2
[a(n-1)+1]/[a(n-2)+1]=(1/4)(n-2)^2
……
[a4+1]/[a3+1]=(1/4)*3^2
[a3+1]/[a2+1]=(1/4)*2^2
[a2+1]/[a1+1]=(1/4)*1^2
两边相乘:
[an+1]/[a1+1]=[(1/4)^(n-1)]*[(n-1)!]^2
an+1=[a1+1][(1/4)^(n-1)]*[(n-1)!]^2
=[(1/2)^(2n-4)]*[(n-1)!]^2
an=-1+[(1/2)^(2n-4)]*[(n-1)!]^2
4.a1=1 an=[(n+1)/n]*a(n+1)
an=[(n+1)/n]*a(n+1)
nan=(n+1)a(n+1)
(n+1)a(n+1)=nan=(n-1)a(n-1)=……=2a2=1a1=1
an=1/n.
a(n+1)=an+5ana(n+1)
1/an=1/a(n+1)+5
1/a(n+1)-1/an=-5
1/an=1/a1-5(n-1)=1/3-5(n-1)=(16-15n)/3
an=3/(16-15n);
2.a1=1 且 a(n+1)=an+2n+1
a(n+1)-an=2n+1
an-a(n-1)=2(n-1)+1
a(n-1)-a(n-2)=2(n-2)+1
……
a3-a2=2*2+1
a2-a1=2*1+1
两边相加:
an-a1=[2(n-1)+1]+[2(n-2)+1]+[2(n-3)+1]+……+[2*2+1]+[2*1+1]
=2[(n-1)+(n-2)+(n-3)+……+2+1]+(n-1)
=n(n-1)+(n-1)
=(n+1)(n-1)
an=a1+(n+1)(n-1)=n^2;
3.a1=1 且 a(n+1)=[(an+1)/4]n^2-1
a(n+1)=[(an+1)/4]n^2-1
a(n+1)+1=[(an+1)/4]n^2
[a(n+1)+1]/(an+1)=(1/4)n^2
[an+1]/[a(n-1)+1]=(1/4)(n-1)^2
[a(n-1)+1]/[a(n-2)+1]=(1/4)(n-2)^2
……
[a4+1]/[a3+1]=(1/4)*3^2
[a3+1]/[a2+1]=(1/4)*2^2
[a2+1]/[a1+1]=(1/4)*1^2
两边相乘:
[an+1]/[a1+1]=[(1/4)^(n-1)]*[(n-1)!]^2
an+1=[a1+1][(1/4)^(n-1)]*[(n-1)!]^2
=[(1/2)^(2n-4)]*[(n-1)!]^2
an=-1+[(1/2)^(2n-4)]*[(n-1)!]^2
4.a1=1 an=[(n+1)/n]*a(n+1)
an=[(n+1)/n]*a(n+1)
nan=(n+1)a(n+1)
(n+1)a(n+1)=nan=(n-1)a(n-1)=……=2a2=1a1=1
an=1/n.
1.a1=3 且 a(n+1)=an+5ana(n+1)
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