发布网友 发布时间:2024-02-23 13:04
共3个回答
热心网友 时间:2024-03-27 01:05
2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1热心网友 时间:2024-03-27 01:06
2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1热心网友 时间:2024-03-27 01:05
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