n
Σ a_k = n^3
k=1
n-1
Σ a_k = (n-1)^3
k=1
n n-1
a_n =Σ a_k -Σ a_k = n^3-(n-1)^3 =3n^2-3n+1
k=1 k=1
2018 1 2018 1 2018 1
Σ ------- = Σ -------------- = Σ -----------
k=2 a_k-1 k=2 3k^2-3k+1-1 k=2 3k*(k-1)
2018 1 1 1 1 1 1
Σ -----------= ----- + ----- + ----- + ----- + .... + -------------
k=2 3k*(k-1) 3*2*1 3*3*2 3*4*3 3*5*4 3*2018*2017
2018 1 1 1 1 1 1 1 1 1 1
Σ -----------= ---*(--- - --- + --- - --- + --- - --- + ... + --- - --- )
k=2 3k*(k-1) 3 1 2 2 3 3 4 2017 2018
2018 1 1 1 1 2017
Σ -----------= ---*(--- - --- ) = -----
k=2 3k*(k-1) 3 1 2018 6054
--
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03/18 14:22,
5年前
, 1F
03/18 14:22, 1F
※ 編輯: a181w (125.227.60.79), 03/18/2019 14:22:34
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03/18 16:58,
5年前
, 2F
03/18 16:58, 2F
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