由a1 = 1,
a1 + 2a2 + 3a3 + ...+ nan = ((n + 1) / 2)a(n + 1) (*)
(*)式取n = 1 得 a2 = 1
當k ≥ 3時
[(*)式取n = k] - [(*)式取n = k - 1] 并將k替換為n 得 nan = [(n + 1)a(n + 1) - nan] / 2
整理得 a(n + 1) / an = 3n / (n + 1)
a(n + 1) = a2 * (a3 / a2) * (a4 / a3) * ...* (a(n + 1) / an)
= (3 * 2 / 3) * (3 * 3 / 4) * ...* (3 * n / (n + 1))
= 2 * 3^(n - 1) / (n + 1)
即
a1 = 1
an = (2 / n) * 3^(n - 2) 當n ≥ 2
應(yīng)該是n*2an吧 否則太難了
設(shè) bn = n*2an
b1 = 2,
bn = 4 * 3^(n - 2) n ≥ 2 是等比數(shù)列
從而 Tn = 2 + 4 * (1 - 3^(n - 1)) / (1 - 3)
= 2 * 3^(n - 1)
恰好對n = 1,2,...成立.