Sequence

👨🏻‍🏫 學術台數學

[匿名]

2014-08-02 7:29 pm

Find the sum in terms of n :
1 + 3^2 + 3^4 + 3^6 + ........ + 3^(4n + 2)

更新1:

Please delete the term 3^4, sorry.

更新2:

To : Mr. Kwok. I know how to handle the '1 ' simply by adding it back to the sum of the rest of the terms, that is easy job, you don't have to worry, just work on the sum of the rest of the terms.

回答 (7)

myisland8132

2014-08-02 11:40 pm

✔ 最佳答案

1 + 3^2 + 3^6 + ... + 3^(4n + 2)

= 1 + 9[1 + 3^4 + ... + 3^(4n)]

= 1 + 9[(3^(4n) - 1)/(3^4 - 1)]

= 1 + (9/80)[(3^(4n) - 1)]

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用戶37586

2014-08-03 6:54 pm

Let S be the sum of 3^2 + 3^6 + ........ + 3^(4n + 2)
so,
S*3^4 = 3^6 + 3^10 + ........ + 3^(4n + 6)
Therefore, the difference is,
80S = 3^(4n + 6) - 3^2
ie. S = [3^(4n + 6) - 9]/80

So, the answer is
1 + [3^(4n + 6) - 9]/80
= [80 + 3^(4n + 6) - 9]/80
= [71 + 3^(4n + 6)]/80

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諸葛諭遜

2014-08-03 5:50 pm

睇唔明係正常的，睇得明係因為佢估倒問者抄錯題目，可能係：

Find the sum in terms of n :
1 + 3^2 + 3^6 + 3^10 + ........ + 3^(4n + 2)

2014-08-03 09:58:51 補充：
如果題目是這樣，那 myisland8132 知識長的答案亦是錯的，他的答案的原題目應該是：

Find the sum in terms of n :
1 + 3^2 + 3^6 + 3^10 + ........ + 3^(4n - 2)

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知足常樂

2014-08-02 11:11 pm

他今早沒有學習我的答案？

https://hk.knowledge.yahoo.com/question/question?qid=7014080200023

2014-08-02 15:13:25 補充：
回少年時老師和郭老師，最後那個 term 未必一定是指 general term。

2014-08-02 15:14:37 補充：
但發問者的補充發問又指出 "Please delete the term 3^4"

所以似乎大家也要先搞清楚該數列的意思才能作答。

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少年時

2014-08-02 10:56 pm

算了吧，郭老師，他不懂你寫的是什麼意思的。
我以下的做法，他也不知錯在那裡。
(錯題目用"錯"方法)
Let S be the sum, that is,
S = 1 + 3^2 + 3^4 + 3^6 + ........ + 3^(4n + 2)
so,
9S = 3^2 + 3^4 + 3^6 + 3^8 + ........ + 3^(4n + 4)
The difference is
8S = 3^(4n + 4) - 1
Therefore, S = [3^(4n + 4) - 1]/8
ie. the sum is [3^(4n + 4) - 1]/8.

2014-08-02 17:02:27 補充：
貓sir 果然心細如絲。��

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Kitty ®

2014-08-02 7:57 pm

Find the sum in terms of n :
1 + 3^2 + 3^4 + 3^6 + ........ + 3^(4n + 2)

= 0 ?

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用戶4366

2014-08-02 7:49 pm

1 + 3^2 + 3^4 + 3^6 + ........ + 3^(4n + 2)

n = 0; (4n + 2) = 2

n= 1; (4n + 2) = 6

3^4 ???????

2014-08-02 11:50:09 補充：
3^4 ???????

n = 1/2

~~~~~~~~~~~~~~

2014-08-02 14:47:07 補充：
1 + 3^2 + 3^4 + 3^6 + ........ + 3^(4n + 2)

3⁰ = 1

(4n + 2) = 0

4n = -2

n = -1/2

2014-08-02 15:09:59 補充：
多謝少年時出手幫助！

~~~~~~~~~~~~~~~~~~

2014-08-02 21:32:16 補充：
所以似乎大家也要先搞清楚該數列的意思才能作答。

正因為睇唔明！

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收錄日期: 2021-04-27 20:37:43
原文連結 [永久失效]:
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