Question

Joanne

a⑩ 怎麼算？

其中是常數,
a
2,又成 - ||
有一數列(a)滿足遞迴關係式
(a = 4 a, = 3
+-cour + ,
14,Fd=(auto} .n22 ' #02*** *
-2
則Q=
Xgk=_
Ao 2 = 1 44-1
azt a = 12 (4+0)
Ap = dati
3+ Q = 2 + a
a = -1 + / a
= -1
a = -2

Answer 1

✨ 最佳解答 ✨

綾

如圖