已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=((n+1)/2)a(n+1)(n∈N*)（1）求an的通项公式 （2）求数列｛n^2an｝的前n项和Tn

Question

已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=((n+1)/2)a(n+1)(n∈N*)
（1）求an的通项公式 （2）求数列｛n^2an｝的前n项和Tn

Answer 1

由a1 = 1,
a1 + 2a2 + 3a3 + ...+ nan = ((n + 1) / 2)a(n + 1) (*)
(*)式取n = 1 得 a2 = 1
当k ≥ 3时
[(*)式取n = k] - [(*)式取n = k - 1] 并将k替换为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) 当n ≥ 2
应该是n*2an吧 否则太难了
设 bn = n*2an
b1 = 2,
bn = 4 * 3^(n - 2) n ≥ 2 是等比数列
从而 Tn = 2 + 4 * (1 - 3^(n - 1)) / (1 - 3)
= 2 * 3^(n - 1)
恰好对n = 1,2,...成立.