已知数列{an}满足a1=4/3,且an+1=4(n+1)an/3an+n
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已知数列{an}满足a1=4/3,且an+1=4(n+1)an/3an+n
![已知数列{an}满足a1=4/3,且an+1=4(n+1)an/3an+n](/uploads/image/z/17127314-26-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%3D4%2F3%2C%E4%B8%94an%2B1%3D4%EF%BC%88n%2B1%EF%BC%89an%2F3an%2Bn)
a(n+1)=4(n+1).an/(3an+n)
3an.a(n+1) + na(n+1) = 4(n+1)an
3 + n/an= 4(n+1)/a(n+1)
4(n+1) [ 1/a(n+1) -1/(n+1) ]= n[ 1/an -(1/n)]
[ 1/a(n+1) -(1/(n+1) ]/[ 1/an -(1/n)] = (1/4)[n/(n+1)]
[ 1/an -(1/n)]/[ 1/a(n-1) -(1/(n-1) ] = (1/4)[(n-1)/n]
[ 1/an -(1/n)]/[ 1/a1 -1/1 ] = (1/4)^(n-1) . (1/n)
1/an -(1/n) = -(1/n).(1/4)^n
1/an = (1/n) [ 1- (1/4)^n ]
an = n/[ 1- (1/4)^n]
再问: 我早就做出来了,不过还是很感谢
3an.a(n+1) + na(n+1) = 4(n+1)an
3 + n/an= 4(n+1)/a(n+1)
4(n+1) [ 1/a(n+1) -1/(n+1) ]= n[ 1/an -(1/n)]
[ 1/a(n+1) -(1/(n+1) ]/[ 1/an -(1/n)] = (1/4)[n/(n+1)]
[ 1/an -(1/n)]/[ 1/a(n-1) -(1/(n-1) ] = (1/4)[(n-1)/n]
[ 1/an -(1/n)]/[ 1/a1 -1/1 ] = (1/4)^(n-1) . (1/n)
1/an -(1/n) = -(1/n).(1/4)^n
1/an = (1/n) [ 1- (1/4)^n ]
an = n/[ 1- (1/4)^n]
再问: 我早就做出来了,不过还是很感谢
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