[問題]費費第一部份-No.29

[Grace]

29、某個數列的通項公式為an+2 = an+an+1(n>=1)，問：a6=?
(1) a6-a5=3;
(2) a7+a8=34;
【答案】：C
【思路】：把(2)化成3a6-2a5=34，然後與(1)聯立，就可以得出結論

請問 a7+a8=34 如何推出 3a6-2a5=34??

Help~ :P

[homaru]

是不是A[sub]n+2[/sub]=A[sub]n[/sub]+A[sub]n+1[/sub]

如果是
那A[sub]7[/sub]+A[sub]8[/sub]
=A[sub]5+2[/sub]+A[sub]6+2[/sub]
=A[sub]5[/sub]+A[sub]6[/sub]+A[sub]6[/sub]+A[sub]7[/sub]
=A[sub]5[/sub]+A[sub]6[/sub]+A[sub]6[/sub]+A[sub]5[/sub]+A[sub]6[/sub]
=2A[sub]5[/sub]+3A[sub]6[/sub]=34

跟他算的3A[sub]6[/sub]-2A[sub]5[/sub]=34不一樣

[huacy]

A[sub]n+2[/sub]=A[sub]n[/sub] + A[sub]n+1[/sub]
A[sub]7[/sub] = A[sub]5[/sub] + A[sub]6[/sub] (n=5)
A[sub]8[/sub] = A[sub]6[/sub] + A[sub]7[/sub] = A[sub]6[/sub] + A[sub]5[/sub] + A[sub]6[/sub] = 2A[sub]6[/sub] + A[sub]5[/sub] (n=6)

所以 A[sub]7[/sub] + A[sub]8[/sub] = 2A[sub]5[/sub] + 3A[sub]6[/sub] = 34

[游客]

謝謝兩位大大的解答......

[Grace]

謝謝兩位高手解惑.....
小妹懂了 :P

[yvetteliao]

對這一題a7+a8=34 如何推出3a6-2a5=34也有疑問
雖然已經有回覆
但不知是不是編輯的問題
我看不出解答的意思ㄟ
請問有人瞭解嗎

A[sub:38c451b353]n+2[/sub:38c451b353]=A[sub:38c451b353]n[/sub:38c451b353] + A[sub:38c451b353]n+1[/sub:38c451b353]
A[sub:38c451b353]7[/sub:38c451b353] = A[sub:38c451b353]5[/sub:38c451b353] + A[sub:38c451b353]6[/sub:38c451b353] (n=5)
A[sub:38c451b353]8[/sub:38c451b353] = A[sub:38c451b353]6[/sub:38c451b353] + A[sub:38c451b353]7[/sub:38c451b353] = A[sub:38c451b353]6[/sub:38c451b353] + A[sub:38c451b353]5[/sub:38c451b353] + A[sub:38c451b353]6[/sub:38c451b353] = 2A[sub:38c451b353]6[/sub:38c451b353] + A[sub:38c451b353]5[/sub:38c451b353] (n=6)

所以 A[sub:38c451b353]7[/sub:38c451b353] + A[sub:38c451b353]8[/sub:38c451b353] = 2A[sub:38c451b353]5[/sub:38c451b353] + 3A[sub:38c451b353]6[/sub:38c451b353] = 34

[dibert8]

a_n+2 = a_n + a_n+1 , 求 a_6
(1) a_6 - a_5 = 3
(2) a_7 + a_8 = 34

(1) , (2) 二個未知數,一個方程式,無限多組解 (不充分)

(2) a_7 + a_8 = 34
=> a_7 + (a_6 + a_7) = 34
=> a_6 + 2*a_7 = 34
=> a_6 + 2*(a_5 + a_6) = 34
=> 2*a_5 + 3*a_6 = 34

(1) & (2) 二個未知數,二個方程式,唯一解 (充分)

[yvetteliao]

謝謝dibert8清楚的解釋
瞭解了