Question

初一数学题 计算:(2^2+1)(2^4+1)(2^8+1)......(2^32+1)

Answer 1

:(2^2+1)(2^4+1)(2^8+1)......(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)/3
=(2^4-1)(2^4+1)(2^8+1)......(2^32+1)/3
=......
=(2^64-1)/3

Answer 2

楼上流了多少汗咱别管他。
解：
(2^2+1)(2^4+1)(2^8+1)......(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^8-1)(2^8+1)......(2^32+1)÷(2^2-1)
=...
=(2^32-1)(2^32+1)÷(2^2-1)
=(2^64-1)/3

2^64等于多少可以不用算出来，老师也不会要求你们算。因为这题的重点不在这里。

追问

为什么要除以2^2-1)

追答

因为前面乘了(2^2-1)

至于前面为什么要乘

好吧，这就是这个题出题的目的。

Answer 3

lz您好
2^2+1 = （2^4 - 1）/ ( 2^2 - 1)
(2^2 + 1)(2^4 + 1) = 2^6 + 2^4 + 2^2 + 1 = (2^8 - 1 )/( 2^2 - 1)
(2^2 + 1)(2^4 + 1)(2^8 + 1) = (2^8 - 1)(2^8 + 1)/(2^2 - 1) = (2^16 - 1)/(2^2 - 1)
依此类推
(2^2+1)(2^4+1)(2^8+1)......(2^32+1) = ( 2^64 - 1) / ( 2^2 - 1)

Answer 4

(2^2+1)(2^4+1)(2^8+1)......(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^8-1)(2^8+1)......(2^32+1)÷(2^2-1)
=(2^32-1)(2^32+1)÷(2^2-1)
=(2^64-1)/3

Answer 5

解：原式 = (2-1)(2+1)(2^2 +1)(2^4 +1)............(2^32 + 1) + 1

= (2^2 - 1 )(2^2 + 1 )(2^4 + 1).........(2^32 + 1) + 1

=(2^4 -1)(2^4 + 1) ........【反复运用平方差公式】

...............

= 2^64 -1 + 1

= 2^64

Answer 6

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追问

怎么了？？？？？？？？？？？？？

追答

太太太太⊙﹏⊙b汗