第1个回答 2009-12-08
下面用a[n]代表数列的项an。
由a[n]=3na[n-1]/(2a[n-1]+ n-1)得
a[n]/n = 3a[n-1]/(2a[n-1]+ n-1),
两边取个倒数,得
n/a[n] = (2/3)*(n-1)/a[n-1] + 2/3
令b[n] = n/a[n],则b[n-1] = (n-1)/a[n-1],上式变为
b[n] = (2/3)b[n-1] + 2/3........(1)
接下来,设
b[n] + t = (2/3)*(b[n-1] + t),即
b[n] = (2/3)b[n-1] + (-1/3)t........(2)
(1)(2)式比较得 t = -2.
就是说,b[n] - 2 = (2/3)*(b[n-1] - 2)。
现在可以看出,c[n] = b[n]-2是一个以2/3为公比的等比数列,且
c[1] = b[1]-2 = 1/a[1]-2 = -4/3.
这样,很容易求得c[n] = -4/3*(2/3)^(n-1).
那么,a[n] = n/(c[n]+2) = n / (2 - 4/3*(2/3)^(n-1)).