一道初中简单数学题啊(50分)

Question

(1+2)*(1+4)*(1+16)*...*(1+2^32)=?
我实在算不出,最多能用完全平方算出,但太麻烦.谁会啊!!
无聊者请别乱吵.

Answer 1

(1+2)*(1+4)*(1+16)*...*(1+2^32)
=-(1-2)*(1+2)*(1+4)*(1+16)*...*(1+2^32)
=-(1-4)*(1+4)*(1+16)*...*(1+2^32)
=-(1-2^64)
=2^64-1

Answer 2

(1+2)*(1+4)*(1+16)*...*(1+2^32)
=(2-1)*(1+2)*(1+4)*(1+16)*...*(1+2^32)
=(4-1)(1+4)*(1+16)*..*(1+2^32)
=(16-1)(1+16)..(1+2^32)
=(2^8-1)(2^8+1)..(1+2^32)
=(2^16-1)(2^16+1)..(1+2^32)
=(2^32-1)(1+2^32)
=2^64-1

Answer 3

(1+2)*(1+4)*(1+16)*...*(1+2^32)
=(2+1)(2^2+1)(2^4+1)...(2^32+1)
=(2-1)(2+1)(2^2+1)(2^4+1)...(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)
=(2^4-1)(2^4+1)...(2^32+1)
...
=2^64-1

Answer 4

(1+2)*(1+4)*(1+16)*...*(1+2^32)=（1-2)(1+2)*(1+4)*(1+16)*...*(1+2^32)/(1-2)
=2^64-1

Answer 5

(1+2)*(1+2+2)*(1+2+2^2).....(1+2+2^31)=(1+2)^32 *2^1 *2^2 *2^3....*2^31=(1+2)^32 *2^2+3+4....31+(1+2)^32 *2^(1+2+3+...31)=(1+2)^32 *2^930
空格键分开了，是不是这答案？？？

Answer 6

原式=（4-1）*(4+1)*(1+16)....(1+2^32）
=（16-1）（16+1）...（1+2^32）
=(2^32-1)(2^32+1)
换算成平方差就可以了