高一数学数列通项的求法

来源：百度知道 编辑：UC知道 时间：2024/09/28 17:48:56
a1=1 an前n项和Sn=n^2*an求an=

(n+1)^2*a(n+1) = S(n+1) = S(n) + a(n+1) = n^2*a(n) + a(n+1),

n(n+2)a(n+1) = n^2a(n),

(n+2)a(n+1) = na(n),

a(n+1)/a(n) = n/(n+2)

a(n)/a(n-1) = (n-1)/(n+1)

...
a(2)/a(1) = 1/3

a(n+1)/a(1) = [n*(n-1)*...*2*1]/[(n+2)*(n+1)*...*4*3]

= 2/[(n+2)(n+1)]

a(n+1) = 2a(1)/[(n+2)(n+1)] = 2/[(n+2)(n+1)],

a(n) = 2/[(n+1)n] = 2[1/n - 1/(n+1)], n = 1,2,...

好像是基本题,最好自己解决,相信我对你有好处.