Question

这道四则运算题怎么做，是不是有简便方法？

笔算的话太麻烦了。。。有没有方法？
(2+1/2)*(2^2+2^-1)*....麻烦大家了！

Answer 1

奥数题，给你解法.给原式乘以（2-1/2），再除以（2-1/2）.从而反复用平方差公式。
原式=（2-1/2）（2+1/2）（2²+1/2²）（2^4+1/2^4）（2^8+1/2^8）（2^16+1/2^16）/(2-1/2)
=(2²-1/2²)（2²+1/2²）（2^4+1/2^4）（2^8+1/2^8）（2^16+1/2^16）/(2-1/2)
=（2^4-1/2^4）（2^4+1/2^4）（2^8+1/2^8）（2^16+1/2^16）/(2-1/2)
=（2^8-1/2^8）（2^8+1/2^8）（2^16+1/2^16）/(2-1/2)
=（2^16-1/2^16）（2^16+1/2^16）/(2-1/2)
=(2^32-1/2^32)/(2-1/2)
=(2^32-1/2^32)x(2/3)
=(2^33-1/2^31)x(1/3)

Answer 2

(2+1/2)(2^2+1/2^)(2^4+1/2^4)(2^8+1/2^8)(2^16+1/2^16)
=(2-1/2)(2+1/2)(2^2+1/2^)(2^4+1/2^4)(2^8+1/2^8)(2^16+1/2^16)/(2-1/2)
=(2^2-1/2^2)(2^2+1/2^)(2^4+1/2^4)(2^8+1/2^8)(2^16+1/2^16)*2/3
=(2^4-1/2^4)(2^4+1/2^4)(2^8+1/2^8)(2^16+1/2^16)*2/3
=(2^8-1/2^8)(2^8+1/2^8)(2^16+1/2^16)*2/3
=(2^16-1/2^16)(2^16+1/2^16)*2/3
=2/3*(2^32-1/2^32)