已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=((n+1)/2)a(n+1)(n∈N*)

来源：学生作业帮编辑：灵鹊做题网作业帮分类：数学作业时间：2024/04/28 19:11:16

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

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.

已知数列{an}中,a1=1,a1+2a2+3a3+…+nan=((n+1)/2)a(n+1)(n∈N*) 设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2) 已知数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2),则a1+a2+a3+…+an=多少? 在数列{an}中,a1+2a2+3a3+.+nan=n(2n+1)(n属于N) 设数列{an}的前n项和为sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*) 已知数列(an)满足a1+2a2+3a3+...+nan=n(n+1)(n+2)求an 已知在数列{an}中,a1=1,nan+1=2(a1+a2+a3+...+an)(n∈N*)(1)求a2,a3,a4（2 数列an满足a1+2a2+3a3+...+nan=(n+1)(n+2) 求通项an 已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1 已知数列{an}中,若a1+2a2+3a3+…+nan=n(n+1)(n+2)则 an= 重点第三小题已知在数列{an}中,a1=1,nan+1=2(a1+a2+a3+...+an)(n∈N*)(1)求a2, 已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N