(a+1)^n=a^n+C(n,1)*a^(n-1)+C(n,2)*a^(n-1)+...........+1 (a-1)^n=a^n-C(n,1)*a^(n-1)+C(n,2)*a^(n-1)-...........+1(or -1) (當n為偶數時末項為1,當n為奇數時末項為-1) (a+1)^n-(a-1)^n=2*C(n,1)*a^(n-1)+2*C(n,3)*a^(n-3)+..... 因此(a+1)^n-(a-1)^n >= 2*C(n,1)*a^(n-1) 即(a+1)^n-(a-1)^n >= 2n*a^(n-1) 設a=2n,則(2n+1)^n-(2n-1)^n >= 2n*(2n)^(n-1) (2n+1)^n-(2n-1)^n > = (2n)^n 亦即(2n+1)^n > = (2n)^n+(2n-1)^n
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