个位数分别是3,9,7,1,3,9,7,1···
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
个位是1有什么不明白可以继续问,随时在线等.