已知数列{an}中,(1)Sn=3+2an,求an(2)a(n+1)=an+2n-1,且a1=1,求an(3)a(n+1
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已知数列{an}中,
(1)Sn=3+2an,求an
(2)a(n+1)=an+2n-1,且a1=1,求an
(3)a(n+1)=a1*2^n,且a1=1,求an
(4)a1=2,a(n+1)=3an+2
下标符号不会打(2)(3)(4)中n+1为a的下标,其余像an,sn,n都为下标,
(1)Sn=3+2an,求an
(2)a(n+1)=an+2n-1,且a1=1,求an
(3)a(n+1)=a1*2^n,且a1=1,求an
(4)a1=2,a(n+1)=3an+2
下标符号不会打(2)(3)(4)中n+1为a的下标,其余像an,sn,n都为下标,
Sn=2an+3
当n≥2时,S(n-1)=2a(n-1)+3
两式相减,得an=2an-2a(n-1),移项得an=2a(n-1)
当n=1时,a1=S1=3+2a1,解出a1= -3
所以an是首项为-3,公比为2的等比数列
an= -3*2^(n-1)
2.设
[a(n+1)+k(n+1)+m]=2(a(n)+kn+m)
2kn-kn=2n
2m-(k+m)=-1
所以k=2,m=1
a(n)+2n+1是等比,q=2 ,a(1)+2+1=4
a(n)+2n+1=4*2^(n-1)
a(n)=4*2^(n-1)-2n-1
3.
a1=1,
a(n+1)=2^n
an=2^n-1
a(n+1)/an=2^n/2^n-1=2
即q=2
an=a1*q^n-1=2^n-1
4.
a(n+1)=3an+2
a(n+1)+1=3an+3
[a(n+1)+1]/[an+1]=3
{an+1}是以3为首项,3为公比的等比数列
an=3^n-1
当n≥2时,S(n-1)=2a(n-1)+3
两式相减,得an=2an-2a(n-1),移项得an=2a(n-1)
当n=1时,a1=S1=3+2a1,解出a1= -3
所以an是首项为-3,公比为2的等比数列
an= -3*2^(n-1)
2.设
[a(n+1)+k(n+1)+m]=2(a(n)+kn+m)
2kn-kn=2n
2m-(k+m)=-1
所以k=2,m=1
a(n)+2n+1是等比,q=2 ,a(1)+2+1=4
a(n)+2n+1=4*2^(n-1)
a(n)=4*2^(n-1)-2n-1
3.
a1=1,
a(n+1)=2^n
an=2^n-1
a(n+1)/an=2^n/2^n-1=2
即q=2
an=a1*q^n-1=2^n-1
4.
a(n+1)=3an+2
a(n+1)+1=3an+3
[a(n+1)+1]/[an+1]=3
{an+1}是以3为首项,3为公比的等比数列
an=3^n-1
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