(2+1)(2的平方+1)(2的4次方+1)......(2的2n次方+1)=?
2个回答
展开全部
原式乘以(2-1)
=(2-1)(2+1)(2^2+1)(2^4+1)……(2^2n+1)+1
=(2^2-1)(2^2+1)(2^4+1)……(2^2n+1)+1
=(2^4-1)(2^4+1)……(2^2n+1)+1
=(2^8-1)……(2^2n+1)+1
|
=2^4n-1+1
=2^4n
=(2-1)(2+1)(2^2+1)(2^4+1)……(2^2n+1)+1
=(2^2-1)(2^2+1)(2^4+1)……(2^2n+1)+1
=(2^4-1)(2^4+1)……(2^2n+1)+1
=(2^8-1)……(2^2n+1)+1
|
=2^4n-1+1
=2^4n
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1
=(2^2-1)(2^2+1)(2^4+1)…(2^2n+1)+1
=.......
=(2^2n-1)(2^2n+1)+1
=2^4n
=(2-1)(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1
=(2^2-1)(2^2+1)(2^4+1)…(2^2n+1)+1
=.......
=(2^2n-1)(2^2n+1)+1
=2^4n
参考资料: http://zhidao.baidu.com/question/68677253.html?si=2
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