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。

Answer 2

1.Sn=a1+2a2+3a3+…+nan=((n+1)/2)a(n+1),S(n-1)=a1+2a2+3a3+…+(n-1)a(n-1)=(n/2)an), 一式减二式得Sn-S(n-1)=nan=((n+1)/2)a(n+1)-(n/2)an), 3nan=(n+1)a(n+1), a(n+1)/an= 3n/(n+1). a(n+1)=a2/a1*a3/a2*a4/a3*┄┄*an/a(n-1)*a(n+1)/an,
=(3*1/2)*(3*2/3)*┄┄*[3*n/(n+1)]. a(n+1)=3^n/(n+1).an的通项公式:an=3^(n-1)/n
2.数列｛n^2an｝的通项公式:an=n*3^(n-1)
前n项和Tn=1*3º+2*3+3*3²+┄┄+n*3^(n-1)
3Tn=1*3+2*3²+3*3³+┄┄+n*3^n，上面一式减二式得
-2Tn=1*3º+1*3+1*3²+┄┄+1*3^(n-1)-n*3^n
-2Tn=(3^n-1)/(3-1)-n*3^n
Tn=[(2n-1)*3^n+1]/4

Answer 3

作差，再化简得a(n+1)/an=n+2/n+1,然后相乘，求出an…