※ 引述《justin0602 (justin)》之銘言：
: http://ppt.cc/8sHW
: 謝謝
: 我對於拆項技巧很不知道怎麼無中生有 才能裂解成兩項
第一題
1/a1 = 1/2 =1-1/a1
故1/a1 + 1/a2 + 1/a3 + ... 1/an
=(1-1/a1) + 1/a2 + 1/a3 + ... +1/an
=1-(1/a1 - 1/a2) + 1/a3 + ... +1/an
=1-(a2-a1)/a1a2 + 1/a3 + ... +1/an
=1- 1/a1a2 + 1/a3 + ... + 1/an
=1- (1/a1a2 - 1/a3) +... + 1/an
=1- (a3-a1a2)/a1a2a3 + ...+ 1/an
=1- 1/a1a2a3 + 1/a4 + ... + 1/an
= ... = 1-1/(a1a2a3...an) < 1
第二題
倒數得1/a_(n+1) = (1+na_n)/a_n = 1/a_n +n
故1/a_2008 = 1/a_2007 +2007
1/a_2007 = 1/a_2006 +2006
.......
1/a_1 = 1/a_0 +0
----------------------------
加總得1/a_2008 = 1/a_0 +(0+1+2+...+2007) =1 + 2007*2008/2 = 2015029
故a_2008= 1/2015029
第三題
由(n+2)/[n!+(n+1)!+(n+2)!] = (n+2)/ n![1+n+1+(n+1)(n+2)]
= (n+2)/ n!*(n+2)^2 = 1/n!(n+2) = (n+1)/ n!(n+1)(n+2) = (n+1)/(n+2)!
= [(n+2)-1]/(n+2)! = (n+2)/(n+2)! - 1/(n+2)! = 1/(n+1)! - 1/(n+2)!
故原式= (1/2!-1/3!) + (1/3!-1/4!) + ... + (1/2003! - 1/2004!)
= 1/2! - 1/2004!
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