Question

计算(2+1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1？

Answer 1

(2+1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=(2-1)(2+1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=(2^2-1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=(2^4-1)(2的4次方+1)······(2的64次方)+1
=...
=2^128-1+1
=2^128,10,2-1)(2+1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=(2^2-1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=(2^4-1)(2的4次方+1)······(2的64次方)+1
=...
=2^128-1+1
=2^128,2,原式乘以(2^2-1)就可以了．2^2-1=3,所以乘了(2^2-1)以后要除以3.然后利用平方差公式:(a+b)*(a-b)=a^2-b^2来计算.
解:原式=（2+1）(2^2-1)(2^2+1))(2^4+1)(2^8+1)(2^16+1)...(2^64+1)/3
=3(2^4-1)(2^4+1)(2^8+1)(2^16+1)...(2^64+1)/3
=3(2^64-1)(2^64+1)/3
=(2^128-1).,1,(2+1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=[（2-1）(2+1)乘以(2平方+1)乘以(2的4次方+1)······(2的64次方)]/(2-1)+1
=(2^2-1)(2平方+1)乘以(2的4次方+1)······(2的64次方)+1
=(2^4-1)(2^4+1)...(2^64+1)+1
=...
=(2^6...,1,(2+1)*(2^2+1)*(2^4+1)······(2^64)+1
=(2-1)(2+1)*(2^2+1)*(2^4+1)······(2^64)+1
=(2^2-1)*(2^2+1)*(2^4+1)······(2^64)+1
=(2^4-1)(2^4+1)······(2^64)+1
=...
=2^128-1+1
=2^128,1,