Sequence

用中學都唔會教到的公式確實不太好。

睇下我點做啦！

Let the name of the sequence 1, - 1, 2, - 2, 3, - 3, 4, - 4, ... be sequence A

很明顯，這條sequence可以睇成由兩條sequence組成的。

我哋首先將sequence A拆開做sequence B和sequence C，其中sequence B取sequence A的第單數項，其中sequence C取sequence A的第雙數項，如下：


sequence B: 1, 2, 3, 4, …

sequence C: - 1, - 2, - 3, - 4, …


不難看出，sequence B和sequence C都是arithmetic sequence。sequence B嘅common difference是1，而sequence C嘅common difference是- 1。


但咁樣都仲未夠，我哋仲要喺sequence B和sequence C嘅項與項之間各加上一項，形成sequence D和sequence E，使sequence A的第單數項是等價於取自sequence D的第單數項，使sequence A的第雙數項是等價於取自sequence E的第雙數項。

其實加上甚麼項都可以，因為我最後會有啲方法整到佢哋無效。不過既然sequence B和sequence C都是arithmetic sequence，而arithmetic sequence嘅general term確實咁容易搵，咁我哋要整到sequence D和sequence E仍然是arithmetic sequence。

整完出嚟嘅sequence D和sequence E係咁樣嘅：

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/sequence_D_sequence_E.jpg

the general term T1(n) of sequence D

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/general_form_1.jpg

the general term T2(n) of sequence E

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/general_form_2.jpg

∴the general term T(n) of sequence A

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/T_1n_T_2n_2.jpg


但這表達式肯定不能令人滿足，所以我哋要進一步簡化至唔駛分case。

現在的要求是：當n是單數，就用T1(n)的值；當n是雙數，就用T2(n)的值。


但是合併case，就要兩個都要同時用，咁點算呢？


我哋可以咁樣解決嘅：無論n是單數還是雙數，T1(n)和T2(n)的值都同時用哂。只不

過當n是單數的時候，我哋就特登做啲手腳，令到T2(n)的值用咗等於冇用；當n是

雙數的時候，我哋就特登做啲手腳，令到T1(n)的值用咗等於冇用。


點樣用咗等於冇用呀？

最直接的方法當然是把T(n)表達至這樣式：
T(n) = T1(n)h1(n) + T2(n)h2(n)

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/h_1n_h_2n.jpg


當然這樣不能直接寫做答案啦！我哋仲要搵埋T1(n)同埋T2(n)嫁！



嚟到呢度，令我諗起(- 1)ⁿ。

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/-1n.jpg


這正正符合T1(n)和T2(n)的要求。

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/h_1n_h_2n_answer.jpg



Hence the general term T(n) of sequence A

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/final2.jpg


圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/result.jpg

The general term of the sequence is the sum of Arithmetic-Geometric series.

a[1] = 1
a[2] = 1+(1+1)(-1) = 1-2 = -1
a[3] = 1+(1+1)(-1) + (1+2)(-1)(-1) = 1-2+3 = 2
and so on

by the definition, check out in search engine:
a=1, d=1, r = -1
a[n]
= [1-(-1)^n]/2 - (1/4) [1-n(-1)^(n-1)+(n-1)(-1)^n], (this from the definition on Net)
Rearrange to (-1)^n
a[n] = 1/4 - (1+2n)(-1)^n / 4
= [1-(2n+1)(-1)^n]/4

You can check it yourself :>