1.已知a1=1,an+1=an+[1/n(n+1)],求an 2.已知a1=1,an+1=3an+1,求an
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a(n+1)=an+[1/n(n+1)]
a(n+1)-an=1/[n(n+1)]=1/n-1/(n+1)
n≥2时,
a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
......................
an-a(n-1)=1/(n-1)-1/n
将上面n-1个等式两边相加
an-a1=1-1/n
an=2-1/n=(2n-1)/n
n=1时上式成立
∴an=(2n-1)/n (n∈N*)
2
∵a(n+1)=3an+1
∴a(n+1)+1/2=3(an+1/2)
∴[a(n+1)+1/2]/(an+1/2)=3
∴{an+1/2}是等比数列,公比为3
又a1+1/2=3/2
∴an+1/2=3/2*3^(n-1)
an=1/2*3^n-1/2=(3^n-1)/2
a(n+1)-an=1/[n(n+1)]=1/n-1/(n+1)
n≥2时,
a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
......................
an-a(n-1)=1/(n-1)-1/n
将上面n-1个等式两边相加
an-a1=1-1/n
an=2-1/n=(2n-1)/n
n=1时上式成立
∴an=(2n-1)/n (n∈N*)
2
∵a(n+1)=3an+1
∴a(n+1)+1/2=3(an+1/2)
∴[a(n+1)+1/2]/(an+1/2)=3
∴{an+1/2}是等比数列,公比为3
又a1+1/2=3/2
∴an+1/2=3/2*3^(n-1)
an=1/2*3^n-1/2=(3^n-1)/2
追问
第2题的3an+1中an为脚标,是3an+(1)
追答
是3an+(1)这样做的
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(1)
a(n+1)=an+ 1/(n(n+1))
=an + 1/n - 1/(n+1)
a(n+1)+1/(n+1) = an+1/n
[a(n+1)+1/(n+1)]/[an+1/n]=1
[an+1/n]/[a1+1]=1
an=2-1/n
(2)
a(n+1)=3an +1
a(n+1)+1/2 = 3(an+1/2)
[a(n+1)+1/2]/[an+1/2]=3
[an+1/2]/[a1+1/2]=3^(n-1)
an= (1/2)3^(n)-1/2
a(n+1)=an+ 1/(n(n+1))
=an + 1/n - 1/(n+1)
a(n+1)+1/(n+1) = an+1/n
[a(n+1)+1/(n+1)]/[an+1/n]=1
[an+1/n]/[a1+1]=1
an=2-1/n
(2)
a(n+1)=3an +1
a(n+1)+1/2 = 3(an+1/2)
[a(n+1)+1/2]/[an+1/2]=3
[an+1/2]/[a1+1/2]=3^(n-1)
an= (1/2)3^(n)-1/2
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1、an+1=an+[1/n(n+1)]
=an+[1/n-1/(n+1)]
∴an+1 +1/(n+1)=an+1/n
数列为等差数列
∴an+1/n=a1+1
∴an=2-1/n
2、设an+1+m=3(an+m),解之得m=1/2
∴数列{an+1/2}为等比数列
∴an+1/2=(a1+1/2)*3^(n-1)
an=(3^n)/2-1/2
=an+[1/n-1/(n+1)]
∴an+1 +1/(n+1)=an+1/n
数列为等差数列
∴an+1/n=a1+1
∴an=2-1/n
2、设an+1+m=3(an+m),解之得m=1/2
∴数列{an+1/2}为等比数列
∴an+1/2=(a1+1/2)*3^(n-1)
an=(3^n)/2-1/2
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