观察下列算式:3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561,...用你所发现的规律求(3-1)(3+1)(3^2+1)(3^4+1)...(3^32+1)+1的个位数是多少.
问题描述：
观察下列算式:3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561,...
用你所发现的规律求(3-1)(3+1)(3^2+1)(3^4+1)...(3^32+1)+1的个位数是多少.
答
因为3^2+1=10，所以(3^0+1)(3^1+1)(3^2+1)(3^3+1)…(3^32+1)的个位为零，再加1，则(3^0+1)(3^1+1)(3^2+1)(3^3+1)…(3^32+1)+1的个位为1
答
1
答
(3-1)(3+1)(3^2+1)(3^4+1)...(3^32+1)+1
=（3^2-1)(3^2+1)(3^4+1)...(3^32+1)+1
=(3^4-1)(3^4+1)...(3^32+1)+1
=...
=3^64
根据上述规律,因为64是4的倍数,3^64个位数与3^4相同,为1