高一数学数列通项的求法a1=1 an前n项和Sn= - 知识问答

高一数学数列通项的求法a1=1 an前n项和Sn=n^2*an求an=

2个回答

(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,...

相关问题