Re: [中學] 數與式 兩題
※ 引述《holysword (= =+)》之銘言：
: 兩題滿有歷史的題目，不過我忘記怎麼做了XD~
: 建議看短網址: http://0rz.tw/m5OGU
: 1.
: 1+2^(-1/2)+3^(-1/2)+...+100^(-1/2)的整數部分為?
: ans:18
: 2.
: (2^(1/3)-1)^(1/3)=a^(1/3)+b^(1/3)+c^(1/3),a,b,c有理數
: 請問a+b+c=?
: ans:1/3
: 想了滿久的，請各位指教~
1.
用小高一的方法的話就是用夾擠
利用2/[√n+√(n+1)] < 1/√n < 2/[√(n-1)+√n]
=>2[√(n+1)-√n] < 1/√n < 2[√n-√(n-1)]
2(√2-√1) < 1 ≦ 1
2(√3-√2) < 1/√2 < 2(√2-√1)
.....
2(√101-√100) < 1/√100 < 2(√100-√99)
加總可得2(√101-1) < 所求 < 19
故可得所求為18.~ 整數部分為18
2.
令 t = 2^(1/3) → t^3 = 2
則(t+1)^3 = t^3 + 3t^2 + 3t + 1 = 2 + 3t^2 + 3t + 1 = 3(t^2 + t + 1)
兩邊同乘t-1 → (t+1)^3 *(t-1) = 3(t^2 + t + 1)*(t-1) = 3(t^3 - 1) = 3
→ t-1 = 3 / (t+1)^3
→ (t-1)^(1/3) = 3^(1/3) / t+1
(2^(1/3)-1)^(1/3) = 3^(1/3) / 2^(1/3)+1
右式分母有理化 分子分母同乘 4^(1/3) - 2^(1/3) + 1
(2^(1/3)-1)^(1/3) = 3^(1/3) * (4^(1/3) - 2^(1/3) + 1) / 2+1
= (4/9)^(1/3) - (2/9)^(1/3) + (1/9)^(1/3)
a+b+c = (4/9)-(2/9)+(1/9) = 1/3
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