Sequence

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)]

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

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)

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。

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。���� ����

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


= 0 ?

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！