qeypour 寫到:(√3+1)^2n
=(4+2√3)^n=(2^n)*(1+√3)^n
=(2^n)*[C(n,0)*(√3)^0+C(n,1)*(√3)+.......+C(n,n)*(√3)^n.....(1)
(√3-1)^2n
=(4-2√3)^n=(2^n)*(1-√3)^n
=(2^n)*[C(n,0)*(-√3)^0+C(n,1)*(-√3)+......+C(n,n)*(-√3)^n...(2)
(1)+(2)得(√3+1)^2n +(√3-1)^2n
=(2^n)*2*[C(n,0)*(√3)^0+C(n,2)*(√3)^2+......+C(n,2k)*(√3)^2k]
=[2^(n+1)]*t,其中t為整數
因為0<(√3-1)^2n<1,所以比(√3+1)^2n大的最小整數就是[2^(n+1)]*t
必為2^(n+1)之倍數,得證
紅色部分稍加修改即可,good job !
(2^n)*(2+√3)^n
=(2^n)*[C(n,0)*2^n+C(n,1)*2^n-1*(√3)+.......+C(n,n)*(√3)^n].....(1)
(√3-1)^2n
=(4-2√3)^n=(2^n)*(2-√3)^n
=(2^n)*[C(n,0)*2^n+C(n,1)*2^n-1*(-√3)+.......+C(n,2k)*(√3)^n]....(2)
(1)+(2)得(√3+1)^2n +(√3-1)^2n
=(2^n)*2*[C(n,0)*2^n+C(n,2)*(√3)^2+......+C(n,2k)*(√3)^2k]