Re: [其他] 遞迴關係式

[vicwk (Victor)]

※ 引述《yuchiao0921 (野格炸彈)》之銘言：
: an=3an-1+2n , a1=3 求an

同除3^n, a(n)/3^n = a(n-1)/3^(n-1) + 2n/3^n
令x(n) = a(n)/3^n, x(1) = 1.
x(n) = x(n-1) + 2n/3^n
x(n-1) = x(n-2) + 2(n-1)/3^(n-1)
.....
x(2) = x(1) + 4/9

則 x(n) = x(1) + 4/9 + 6/27 + ... + 2(n-1)/3^(n-1) + 2n/3^n
x(n)/3 = x(1)/3  +     4/27 + ... + 2(n-2)/3^(n-1) + 2(n-1)/3^n + 2n/3^(n+1)
2x(n)/3 = 2x(1)/3 + 4/9 + 2/27 + ... + 2/3^(n-1) + 2/3^n - 2n/3^(n+1)
= 3(1 - 1/3^(n+1)) - 16/9 - 2n/3^(n+1)
= 11/9 - (2n+3)/3^(n+1)

x(n) = 11/6 - (n + 3/2)/3^n, a(n) = (11/2)*3^(n-1) - (n + 3/2).

代回原式驗算
a(n) = (11/2)*3^(n-1) - (n + 3/2)
a(n-1) = (11/2)*3^(n-2) - (n + 1/2)
a(n) - 3a(n-1) = (11/2)*3^(n-1) - (n + 3/2) - (11/2)*3^(n-1) + 3(n + 1/2)
= 2n.

--
※ 發信站: 批踢踢實業坊(ptt.cc) 
◆ From: 163.22.20.4
推 yuchiao0921 :謝謝