（2+1）（2的2次方+1）（2的4次方+1)(2的8次方)…（2的64次方）
1个回答
展开全部
(2-1)(2+1)（2^2-1）(2^2+1)(2^4+1)...(2^64+1)
=（2^2-1）[(2-1)(2+1)（2^2-1）(2^2+1)(2^4+1)...(2^64+1)]
=3(2-1)(2+1)（2^2-1）(2^2+1)(2^4+1)...(2^64+1)
=3(2^2-1)(2^2+1)(2^4+1)...(2^64+1)
=3(2^4-1)(2^4+1)...(2^64+1)
=3(2^8-1)...(2^64+1)
...
=3(2^64-1)(2^64+1)
=3(2^128-1)
=3×2^128-3
=（2^2-1）[(2-1)(2+1)（2^2-1）(2^2+1)(2^4+1)...(2^64+1)]
=3(2-1)(2+1)（2^2-1）(2^2+1)(2^4+1)...(2^64+1)
=3(2^2-1)(2^2+1)(2^4+1)...(2^64+1)
=3(2^4-1)(2^4+1)...(2^64+1)
=3(2^8-1)...(2^64+1)
...
=3(2^64-1)(2^64+1)
=3(2^128-1)
=3×2^128-3
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
