已知数列{an}满足(n+1)an+1=an+n,且a1=2,则a2010=?

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已知数列{an}满足(n+1)an+1=an+n,且a1=2,则a2010=?
已知数列{an}满足(n+1)an+1=an+n,且a1=2,则a2010=?

已知数列{an}满足(n+1)an+1=an+n,且a1=2,则a2010=?
因为(n+1)a(n+1)=an+n,
所以a(n+1)=(an+n)/(n+1),
a(n+1)-1=(an+n)/(n+1)-1,
即a(n+1)-1=(an-1)/(n+1),
取倒数可得：
1/[ a(n+1)-1]= (n+1) /(an-1),
∴ [ a(n+1)-1] /(an-1) = 1/(n+1).
所以
(a2-1)/(a1-1)=1/2,
(a3-1)/(a2-1)=1/3,
(a4-1)/(a3-1)=1/4,
…………
(an-1) /(a(n-1)-1) = 1/n.
以上各式相乘得：
(an-1) /(a1-1)=1/(2*3*4*……*n),
a1=2代入上式可得：
an-1=1/(n!),
an=1+1/(n!).
∴a2010=1+1/2010!.

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