有數題不太明白如何做,麻煩詳細步驟
1.2^n+1 / 2^n-1
2.3^n+1 x 3^n-2 / 3^2n+1
3.5x2^n+2 - 2^n+1 / 2^n+3+2^n
關於指數定律公式問題
2007-02-18 3:05 am
回答 (2)
2007-02-18 4:44 am
參考: king
2007-02-18 3:22 am
1.
2^(n + 1) / 2^(n - 1)
= 2^[(n + 1) - (n - 1)]
= 2^(n + 1 - n + 1)
= 2^2
= 4
2.
3^(n + 1) x 3^(n - 2) / 3^(2n + 1)
= 3^[(n + 1) + (n - 2) - (2n + 1)]
= 3^(n + 1 + n - 2 - 2n - 1)
= 3^(-2)
= 1/9
3.
[5 x 2^(n + 2) - 2^(n + 1)] / [2^(n + 3) + 2^n]
= 2^(n + 1) x (5 x 2 - 1) / 2^n x (2^3 + 1)
= 2 x 9 / 9
= 2
2^(n + 1) / 2^(n - 1)
= 2^[(n + 1) - (n - 1)]
= 2^(n + 1 - n + 1)
= 2^2
= 4
2.
3^(n + 1) x 3^(n - 2) / 3^(2n + 1)
= 3^[(n + 1) + (n - 2) - (2n + 1)]
= 3^(n + 1 + n - 2 - 2n - 1)
= 3^(-2)
= 1/9
3.
[5 x 2^(n + 2) - 2^(n + 1)] / [2^(n + 3) + 2^n]
= 2^(n + 1) x (5 x 2 - 1) / 2^n x (2^3 + 1)
= 2 x 9 / 9
= 2
收錄日期: 2021-04-12 23:23:01
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https://hk.answers.yahoo.com/question/index?qid=20070217000051KK03020