f3指數定律

2008-11-04 6:10 am

[4(n+1)-4(n-1)]/4(n)

p.s.(n)n=次方

回答 (4)

2008-11-05 4:59 am

✔ 最佳答案

[4(n+1)-4(n-1)]/4(n) = [4^(n+1)-4^(n-1)]/4^n..............4^a=4既a次方

= [4(4^n) -(4^n)(-4)]/4^n
=[4(4^n) +4(4^n)]/4^n
=8(4^n)/4^n
=8

參考: 會考數學拿c既人~

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2008-11-06 12:21 am

[4^(n+1)-4^(n-1)]/4^(n)
=4^(n+1-n)-4^(n-1-n)
=4-(1/4)
=15/4

參考: myself

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2008-11-04 2:15 pm

1)

[4^(n + 1) - 4^(n - 1)] / 4^n

= (4^n * 4 - 4^n * 4) / 4^n

= 0 / 4^n

= 0

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2008-11-04 6:50 am

我讀文科的，不過嘗試作答。

[4(n+1)-4(n-1)]/4(n)
= [4(n+1) / 4(n)] - [4(n-1) / 4(n)]
= 4(n+1-n) - 4(n-1-n)
= 4(1) - 4(-1)
= 4 - 1/4
= 3.75 (即3又4分3)

參考: 自己

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