已知数列｛an｝满足a1=4/3，且an+1=4（n+1）an/3an+n
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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/野友孝an -(1/n)]/颂稿[ 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/告旁n) = -(1/n).(1/4)^n
1/an = (1/n) [ 1- (1/4)^n ]
an = n/[ 1- (1/4)^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/野友孝an -(1/n)]/颂稿[ 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/告旁n) = -(1/n).(1/4)^n
1/an = (1/n) [ 1- (1/4)^n ]
an = n/[ 1- (1/4)^n]
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我早就做出来了，不过还是很感谢
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