※ 引述《mj813 (薩坨十二惡皆空)》之銘言：
: 設 f(x)＝16^x / (4+16^x)
: 求 f(1/6)+f(2/6)+f(3/6)+f(4/6)+f(5/6)
: 拜託解惑，感激各位前輩！
16^x = 2^4x
f(x) = 2^4x/(4+2^4x) = 2^6x/(2^(2x+2)+2^6x)
f(n/6) = 2^n/(2^(2+n/3)+2^n)
n f(n/6)
1 2/(2^(2+1/3)+2)=1/[2^(4/3)+1]
2 4/[2^(2+2/3)+4]=1/[2^(2/3)+1]
3 8/[2^(2+1)+8]=1/2
4 16/[2^(2+4/3)+16]=1/[2^(-2/3)+1]=2^(2/3)/[1+2^(2/3)]
5 32/[2^(2+5/3)+32]=1/[2^(-4/3)+1]=2^(4/3)/[1+2^(4/3)]
f(1/6)+f(5/6) = f(2/6)+f(4/6) = 1
Ans = 5/2
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※ 編輯: deathcustom (211.23.191.211 臺灣), 10/09/2024 14:02:27