精華區beta GMAT 關於我們 聯絡資訊
※ 引述《chilis (龐克新生報到^_^)》之銘言: : Q2: : For which of the following functions f is f(x) = f(1-x) for all x? : A. f(x) = 1 - x : B. f(x) = 1 - x2 : C. f(x) = x2 - (1-x)2 : D. f(x) = x2(1-x)2 : E. f(x) = x / (1-x) : ---------------------------------------------------------------------------------------------------------------------- : Answer: D 不懂題目?? : Q9: : A certain computer program generates a sequence of numbers a1, a2, … , : a n such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that : 3 ? k ? n. If n > 6, then a7 = ? : A. 32 : B. 43 : C. 64 : D. 100 : E. 128 : ---------------------------------------------------------------------------------------------------------------------- : Answer: B : a1 =a0+2a-1=1 : a2 =a1+2a0= 1 : a3=a2+2a1=1+2*1=3=2^1+1 : a4=a3+2a2=3+2*1=5=2^2+1 : a5=a4+2a3=5+2*3=11 : a6=a5+2a4=11+2*5=21 : a7=a6+2a5=21+2*11=43 : 找不出規則變化,有快速解法嗎? : 感謝:) 經過複雜的運算後,有導出公式如下: An= ( 2^(n-1) * (-3 + (-1)^(n-1) ) -1 ) / (-3) 有點給他複雜,而且考試時間那麼趕,應該沒時間讓你導公式, 所以直接土法煉鋼從A1算到A7比較快XDD 不過土法煉鋼比較快的做法就是: An會剛好=2*An-1 + (-1)^(n-1) 也就是會剛好=前一個數的2倍,再+1或-1 (n是偶數就-1,n是奇數就+1) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 118.166.135.134