Question

(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)

Answer 1

因为（2-1）=1
所以原式=（2-1）(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
= (2²-1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=……
=(2的n次方-1)(2的n次方+1)
=2的n+1次方-1

Answer 2

题目应改为：(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)【n=2^m，m为自然数】(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=（2-1）(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=(2²-1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=2的2n次方-1

Answer 3

(2+1)(2²+1)(2⁴+1)(2^8+1)……(2ⁿ+1)
=(2-1)(2+1)(2²+1)(2⁴+1)(2^8+1)……(2ⁿ+1)
=(2²-1)(2²+1)(2⁴+1)(2^8+1)....(2ⁿ+1)
=(2⁴-1)(2⁴+1)(2^8+1)....(2ⁿ+1)
=..................
=(2ⁿ-1)(2ⁿ+1)=2²ⁿ-1

Answer 4

(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=（2-1）(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=(2²-1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=2的n次方-1

Answer 5

(2+1)(2²+1)(2的四次方+1)(2的八次方+1)……(2的n次方+1)
=（2^-1)(2^2+1)(2^4+1)(2^8+1)......(2^n+1)
=(2^4-1)(2^4+1)(2^8+1)......(2^n+1)
=(2^8-1)(2^8+1)......(2^n+1)
=2^（2n）-1