计算(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1计算(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1注：上式中第二个括号中的2是上标,后面括号中的：4、8、16、32都是上标

(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)+1
=(2^2-1)*(2^2+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
如有雷同···

几年级作业这么难啊？

平方差
因为2-1=1
所以(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)+1
=(24-1)(24+1)(28+1)(216+1)(232+1)+1
反复用
=264-1+1
=2的64次方

(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1
=(2²-1)(22+1)(24+1)(28+1)(216+1)(232+1)+1
=。。。。
=(232-1)(232+1)+1
=264-1+1
=2的64次方

把这个式子最左边乘以一个(2-1)不改变结果,然后用平方差公式就能做出来了,结果是2的64次方