作业帮计算：2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/16 06:37:53
计算：2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+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).(3^32+1)+1 关键是:把2变成3-1
=(3^2-1)(3^2+1)(3^4+1).(3^32+1)+1能后多次运用平方差公式
=(3^4-1)(3^4+1).(3^32+1)+1
=...
=(3^32-1)(3^32+1)+1
=3^64-1+1
=3^64
= 3.4336838202925 * 10^30
=(3-1)(3+1)(3^2+1)(3^4+1).(3^32+1)+1 关键是:把2变成3-1
=(3^2-1)(3^2+1)(3^4+1).(3^32+1)+1能后多次运用平方差公式
=(3^4-1)(3^4+1).(3^32+1)+1
=...
=(3^32-1)(3^32+1)+1
=3^64-1+1
=3^64
= 3.4336838202925 * 10^30