求教(2+1）(22+1）(24+1）(28+1）…(264+1）+1等于多少?
问题描述：
求教(2+1）(22+1）(24+1）(28+1）…(264+1）+1等于多少?
答
原式=(2-1)(2+1)(2^2+1)(2^4+1)( 2^8+1)( 2^16+1)( 2^32+1)( 2^64+1)+1
连续利用平方差公式可得=2^128-1+1=2^128
即
（2+1）(2^2+1)(2^4+1)......(2^64+1) +1
=(2-1)（2+1）(2^2+1)(2^4+1)......(2^64+1) +1
=(2^2-1)(2^2+1)(2^4+1)......(2^64+1) +1
=(2^4-1)(2^4+1)......(2^64+1) +1
=2^128-1 +1
=2^128
答
(2+1)(2^2+1)(2^4+1)(2^8+1）…(2^64+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1）…(2^64+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1）…(2^64+1)+1
=(2^4-1)(2^4+1)(2^8+1）…(2^64+1)+1
=(2^8-1)(2^8+1）…(2^64+1)+1
=(2^16-1).(2^64+1)+1
.
=(2^64-1)(2^64+1)+1
=2^128-1+1
=2^128
PS：2^2表示2的2次方,.2^128表示2的128次方.