已知数列{an}满足(n+1)an+1=an+n,且a1=2,则a2010=?
问题描述:
已知数列{an}满足(n+1)an+1=an+n,且a1=2,则a2010=?
答
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-...