费马数Fn=2^(2^n),求证(F0)(F1)...*(Fn-1)+2=Fn.(n>=2)

clement19862022-10-04 11:39:542条回答

费马数Fn=2^(2^n),求证(F0)*(F1)*...*(Fn-1)+2=Fn.(n>=2)
打错了，Fn=2^(2^n)+1

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fc453b446f0b7787 共回答了14个问题 | 采纳率100%: 解题须知:(a + b)(a - b) = a^2 - b^2(2^a + 1)(2^a - 1) = (2^a)^2 - 1 = 2^2a - 1(2^(2^a) + 1)(2^(2^a) - 1) = 2^(2 * 2^a) - 1 = 2^(2^(a +1)) - 12^(2^0) - 1 = 1(F0)(F1)...(Fn-1) + 2= (1)(F0)(F1)...(Fn-1) ...; 1年前

wodedaoyu 共回答了8个问题 | 采纳率: Fn-2=2^(2^n)-1=(2^(2^(n-1)+1)*(2^(2^(n-1)-1)=(Fn-1)*(2^(2^(n-2)+1)*(2^(2^(n-2)-1)=(Fn-1)*(Fn-2)*...........*F0*(2-1)=)=(Fn-1)*(2^(2^(n-2)+1)*(2^(2^(n-2)-1)=(Fn-1)*(Fn-2)*...........*F0
将-2右移得到：Fn=)=)=(Fn-1)*(2^(2^(n-2)+1)*(2^(2^(n-2)-1)=(Fn-1)*(Fn-2)*...........*F0+2得证
也可以用数学归纳法。; 1年前

费马数Fn=2^(2^n),求证(F0)*(F1)*...*(Fn-1)+2=Fn.(n>=2)