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更新1:
ANS: 1/{(k^2 +1)(k^2 +2)(k^2 +3)...(k^2 +2k)}
數學!factional問題
回答 (2)
2011-02-08 5:39 am
✔ 最佳答案
{[(k+1)！]^2 (k^2)！} / {(k!)^2 [(k+1)^2]！}=??Sol{[(k+1)！]^2 (k^2)！} / {(k！)^2 [(k+1)^2]！}={[(k+1)！]^2/ (k！)^2}*{(k^2)!/[(k+1)^2]！}={k！*k！*(k+!)*(k+1)/(k！)^2}*{(k^2)！/[k！*k！*(k+1)*(k+1)]}=(k+!)*(k+1)*{1/[(k+1)*(k+1)]}=1 
2011-02-08 7:26 am
[(k+1)!]² (k²)!
-------------------
(k!)² [(k+1)²]!
=
(k!)² (k+1)² (k²)!
--------------------------
(k!)² (k² + 2k + 1)!
=
2011-02-07 23:26:14 補充：
(k+1)² (k²)!
---------------------------------------------
(k²)! (k²+1) (k²+2) ... (k²+2k+1)
=
(k²+2k+1) (k²)!
-----------------------------------------------------
(k²)! (k²+1) (k²+2) ... (k²+2k)(k²+2k+1)
=
1 / [(k^2 +1)(k^2 +2)(k^2 +3)...(k^2 +2k)]
-------------------
(k!)² [(k+1)²]!
=
(k!)² (k+1)² (k²)!
--------------------------
(k!)² (k² + 2k + 1)!
=
2011-02-07 23:26:14 補充：
(k+1)² (k²)!
---------------------------------------------
(k²)! (k²+1) (k²+2) ... (k²+2k+1)
=
(k²+2k+1) (k²)!
-----------------------------------------------------
(k²)! (k²+1) (k²+2) ... (k²+2k)(k²+2k+1)
=
1 / [(k^2 +1)(k^2 +2)(k^2 +3)...(k^2 +2k)]
收錄日期: 2021-04-21 22:18:59
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https://hk.answers.yahoo.com/question/index?qid=20110207000051KK01424