Question

计算2^3(3^2+1)(3^4+1）（3^8+1)....(3^32+1)+1，求详细过程

Answer 1

提示：2³=3²-1，然后反复运用平方差公式。
2³×(3²+1)(3⁴+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3²-1)(3²+1)(3⁴+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3⁴-1)(3⁴+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3^8-1)(3^8+1)(3^16+1)(3^32+1)+1
=(3^16-1)(3^16+1)(3^32+1)+1
=(3^32-1)(3^32+1)+1
=3^64-1+1
=3^64

3^64表示3的64次方。

Answer 2

2^3(3^2+1)(3^4+1）（3^8+1)....(3^32+1)+1
=8*(3^2+1)(3^4+1）（3^8+1)....(3^32+1)+1
=(3^2-1)*(3^2+1)(3^4+1）（3^8+1)....(3^32+1)+1
=(3^4-1)(3^4+1）（3^8+1)....(3^32+1)+1
=(3^8-1)(3^8+1)....(3^32+1)+1
=(3^16-1)....(3^32+1)+1
.......
=(3^32-1)(3^32+1)+1
=3^64-1+1
=3^64

Answer 3

将2^3化成(3^2-1),
再逐次利用平方差公式，即有：
原式=(3^4-1)(3^4+1)(3^8+1)...(3^32+1)
=(3^8-1)(3^8+1)......(3^32+1)
=....
=3^64-1