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计算:(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1)...({2}^{256}+1)
人气:106 ℃ 时间:2020-03-22 04:10:53
解答
答:补上因式(2-1)后重复利用平方差公式
(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1)...({2}^{256}+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1).(2^256+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1).(2^256+1)
=(2^4-1)(2^4+1)(2^8+1).(2^256+1)
=(2^8-1)(2^8+1).(2^256+1)
=(2^16-1).(2^256+1)
=2^512 -1
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