己知f(x)=2(5^x),若f(k+1)-f(k)=1000,k值是?
要有步驟,thz:)
一條maths
2012-02-05 2:39 am
回答 (4)
2012-02-05 2:48 am
✔ 最佳答案
f(k+1) - f(k) = 10002(5ᵏ⁺¹) - 2(5ᵏ) = 10005ᵏ⁺¹ - 5ᵏ = = 5005ᵏ (5 - 1) = 5005ᵏ = 500/4 = 125 = 5³ k = 3 
2012-02-06 12:24 am
f(x)=2(5^x)
f(k+1)=2[5^(k+1)]
f(k)=2(5^k)
f(k+1)-f(k)=1000
2[5^(k+1)]-2(5^k)=1000
5^(k+1)-5^k=500
(5^k)(5)-5^k=500
5^k(5-1)=500
5^k=125
5^k=5^3
k=3
f(k+1)=2[5^(k+1)]
f(k)=2(5^k)
f(k+1)-f(k)=1000
2[5^(k+1)]-2(5^k)=1000
5^(k+1)-5^k=500
(5^k)(5)-5^k=500
5^k(5-1)=500
5^k=125
5^k=5^3
k=3
2012-02-05 5:20 am
f(k + 1) - f(k) = 1000
2[5^(k + 1)] - 2(5^k) = 1000
5^(k + 1) - (5^k) = 500
(5^k)(5 - 1) = 500
5^k = 125
k = 3
2[5^(k + 1)] - 2(5^k) = 1000
5^(k + 1) - (5^k) = 500
(5^k)(5 - 1) = 500
5^k = 125
k = 3
2012-02-05 2:51 am
f(k+1) - f(k) = 10002(5ᵏ⁺¹) - 2(5ᵏ) = 10005ᵏ⁺¹ - 5ᵏ = = 5005ᵏ (5 - 1) = 5005ᵏ = 500/4 = 125 = 5³ k = 3不是5年級教的嗎?(I am 5年級)
參考: k = 3
收錄日期: 2021-04-21 14:41:09
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https://hk.answers.yahoo.com/question/index?qid=20120204000051KK00697