[分享] 遞迴數列自創解法
沒想過要當補教名師
某天家教沒帶書忘了倍數遞迴怎麼解
卻反而偶然自創好用的新解法 某些情況(如公比為整數)解比較快
公比為分數一樣可以解
希望它有機會傳下去囉
或有人能幫我探索此解法的限制
例題： a(1)=4 a(n)=2/3 a(n-1)+5 求a(n)之通解？
1.原標準解法：
a(1)=4;a(2)= 23/3
a(2)-a(1) = 11/3
a(3)-a(2) = 2/3[a(2)-a(1)]
a(4)-a(3) = 2/3[a(3)-a(2)] = (2/3)^2[a(2)-a(1)]
a(5)-a(4) = 2/3[a(4)-a(3)] = (2/3)^3[a(2)-a(1)]
:
:
:
a(n)-a(n-1)= 2/3[a(n-1)-a(n-2)] = (2/3)^(n-2)[a(2)-a(1)]
___________________________________________________________
a(n)-a(1) = {1+2/3+(2/3)^2+(2/3)^3+……+(2/3)^(n-2)}x[a(2)-a(1)]
= {1[1-(2/3)^(n-1)]}/(1-2/3)x[a(2)-a(1)]
=> a(n) = 15 - 11(2/3)^(n-1)
2.妞式解法
a(1)=4 ; a(2)=23/3 ; a(3)=91/9
d(1)=a(2)-a(1)= 11/3
d(2)=a(3)-a(2)= 22/9
=>d(n)=11/3 x (2/3)^(n-1)
<以下重點>
令a(n) = Ad(n)+B
代入a(1)=4 = 11/3A + B
a(2)=23/3 = 22/9A + B
聯立解得 11/9 A = -11/3 =>A = -3
B = 15
代回a(n)解得 a(n) = 15-11 (2/3)^(n-1)
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※ 編輯: njunju 來自: 114.40.101.132 (03/07 18:28)
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