230问答网 > 观察算式2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1不用计算器能算出来吗？结果个位数字是几？

观察算式2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1不用计算器能算出来吗？结果个位数字是几？

2024-12-17 01:15:49

推荐回答（3个）

回答1：

2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3-1)*(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3^8-1)(3^8+1)(3^16+1)(3^32+1)+1
=(3^16-1)(3^16+1)(3^32+1)+1
=(3^32-1)(3^32+1)+1
=(3^64-1) +1=3^643^64的个位数字是3,9,7,1的1,所以3^64的个位数字是1

回答2：

2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
=3^64-1+1
=3^64
=(3^4)^16
=81^16
个位是1

回答3：

解： 2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1 ={2[3+1)(3-1)](3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1}/(3-1) ={2[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^16+1)(3^32+1)+1}/(3-1) ={2[(3^4-1)(3^4+1)](3^8+1)(3^16+1)(3^32+1)+1}/(3-1) ={2[(3^8-1)(3^8+1)](3^16+1)(3^32+1)+1}/(3-1) ={2[(3^16-1)(3^16+1)](3^32+1)+1}/(3-1) ={2[(3^32-1)(3^32+1)]+1}/(3-1) ={2(3^64-1)+1}/(3-1) =(3^64-1)+1 =3^64