Anonymous
Anonymous asked in 科學及數學數學 · 6 years ago

# Sequence

Find the sum in terms of n :

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

Update:

Please delete the term 3^4, sorry.

Update 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.

### 7 Answers

Rating
• 6 years ago
Favorite Answer

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)]

• Login to reply the answers
• 6 years ago

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

• Login to reply the answers
• 6 years ago

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)

• Login to reply the answers
• 6 years ago

他今早沒有學習我的答案？

https://hk.knowledge.yahoo.com/question/question?q...

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。

• Login to reply the answers
• How do you think about the answers? You can sign in to vote the answer.
• 6 years ago

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。���� ����

• Login to reply the answers
• 6 years ago

Find the sum in terms of n :

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

= 0 ?

• Login to reply the answers
• CK
Lv 7
6 years ago

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！

• Login to reply the answers
Still have questions? Get your answers by asking now.