观察下列各式: (x-1)(x+1)=x2-1, (x-1)(x2+x+1)=x3-1, (x-1)(x3+x2+x+1)=x4-1, (x-1)(x4+x3+x2+x+1)=x5-1, (1)根据前面各式的规律可得:(x-1)(xn+xn
问题描述:
观察下列各式:
(x-1)(x+1)=x2-1,
(x-1)(x2+x+1)=x3-1,
(x-1)(x3+x2+x+1)=x4-1,
(x-1)(x4+x3+x2+x+1)=x5-1,
(1)根据前面各式的规律可得:(x-1)(xn+xn-1+…+x2+x+1)=______(其中n为正整数).
(2)根据(1)求1+2+22+23+…+262+263的值,并求出它的个位数字.
答
(1)根据各式的规律可得:(x-1)(xn+xn-1+…+x2+x+1)=xn+1-1;(2)根据各式的规律得:1+2+22+23+…+262+263=(2-1)(263+262+…+23+22+2+1)=264-1,∵21=2,22=4,23=8,24=16,25=32,…,且64÷4=16,∴264...