Question

计算：3（2²+1）（2^4+1）(2^8+1)(2^16+1)

Answer 1

提示你 将3写成2的平方-1

然后连续平方差公式

我想你那么聪明 应该会了

Answer 2

3（2²+1）（2^4+1）(2^8+1)(2^16+1)
=（2^2-1）（2²+1）（2^4+1）(2^8+1)(2^16+1)
=（2^4-1）（2^4+1）(2^8+1)(2^16+1)
=（2^8-1）（2^8+1)(2^16+1)
=（2^16-1）（2^16+1)
=2^32-1

Answer 3

3(2^2+1)(2^4+1)(2^8+1)(2^16+1
=3(2²-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2²-1)
=3(2^4-1)(2^4+1)(2^8+1)(2^16+1)/(2²-1)
=3(2^8-1)(2^8+1)(2^16+1)/(2²-1)
=3(2^16-1)(2^16+1)/(2²-1)
=3(2^64-1)/3
=2^64-1

Answer 4

(2ˆ2-1)(2ˆ2+1)(2ˆ4+1)(2ˆ8+1)(2ˆ16+1)
=(2ˆ4-1)(2ˆ4+1)(2ˆ8+1)(2ˆ16+1)
=(2ˆ8-1)(2ˆ8+1)(2ˆ16+1)
(2ˆ16-1)(2ˆ16+1)
=2ˆ32-1

可追问啊