(2+1)(2的平方加1)(2的4次方加1).(2的32次方加一)加1=?
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(2+1)(2的平方加1)(2的4次方加1).(2的32次方加一)加1=?
(2+1)(2的平方加1)(2的4次方加1).(2的32次方加一)加1=?
(2+1)(2的平方加1)(2的4次方加1).(2的32次方加一)加1=?
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=1×(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2²-1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^4-1))(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^16-1)(2^16+1)(2^32+1)+1
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64
原式=(2-1)(2+1)(2^2 +1)(2^4 +1)……(2^32 +1) +1
=(2^2 -1)(2^2 +1)(2^4 +1)……(2^32 +1) +1
=……
=(2^64 -1) +1
=2^64
思路:添一项 (2-1) 连续使用平方差公式
=(2-1)(2+1)(2^2 +1)(2^4 +1)……(2^32 +1) +1
=(2^2 -1)(2^2 +1)(2^4 +1)……(2^32 +1) +1
=……
=(2^64 -1) +1
=2^64