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※ 引述《firzen11589 (逆天)》之銘言：
: 後天要資工筆試 請問一些考古題
: 1 1 1
: Z+ --- = 1 求 Z^2010 + -------- = ?
: z Z^2010
: 2 lim [(x^3 + 3x^2 + 5)^(1/3) - (x^5 + 2x^4 + 7x^3 + x +3)^(1/5)
: x→∞
: 3 f(0,y) = y+1 , f(x+1,0) = f(x,1) , f(x+1,y+1) = f(x,f(x+1,y)),那麼f(5,0) = ?
f(0,0) = 1, f(0,1) = 2, f(0,2) = 3, f(0,3) = 4 , f(0,4) = 5, f(0,5) = 6
f(1,0) = f(0+1,0) = f(0,1) = 2
f(1,1) = f(0+1,0+1) = f(0,f(1,0)) = f(0,2) = 3
f(1,2) = f(0+1,1+1) = f(0,f(1,1)) = f(0,3) = 4
f(1,n) = f(0+1,n-1+1) = f(0,f(1,n-1)) = f(1,n-1) +1
=> f(1,n) = n+f(1,0) = n+2
f(2,0) = f(1+1,0) = f(1,1) = 3
f(2,1) = f(1+1,0+1) = f(1,f(2,0)) = f(1,3) = 5
f(2,n) = f(1+1,n-1+1) = f(1,f(2,n-1)) = f(2,n-1)+2
=> f(2,n) = 2n+f(2,0) = 2n+3
f(3,0) = f(2+1,0) = f(2,1) = 5
f(3,1) = f(2+1,0+1) = f(2,f(3,0)) = 13
f(3,n) = f(2+1,n-1+1) = f(2,f(3,n-1)) = 2*f(3,n-1) +3
=> f(3,n) = a*2^n + b , 代入 f(3,0) = 5, f(3,1) = 13
=> a+b = 5, 2a+b = 13, a=8, b=-3
=> f(3,n) = 8*2^n -3
f(4,0) = f(3+1,0) = f(3,1) = 13
f(4,1) = f(3+1,0+1) = f(3,f(4,0)) = f(3,13) = 8*2^13 -3
f(5,0) = f(4+1,0) = f(4,1) = 8*2^13-3 = 65533
: 10 k
: 4{x|1≦Σ ------ ≦2}為若干個區間的聯集,試問區間的長度為?
: k=1 (x-k)
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