51问答网 > 计算2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1，急急急急

计算2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1，急急急急

2024-12-20 07:36:13

推荐回答（2个）

回答1：

解答：
2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1

=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
连续利用平方差公式
=3^64-1+1
=3^64

回答2：

2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
=(3-1)(3+1)(3^2+1)(3^4+1)......(3^32+1)+1 关键是:把2变成3-1
=(3^2-1)(3^2+1)(3^4+1).....(3^32+1)+1能后多次运用平方差公式
=(3^4-1)(3^4+1)......(3^32+1)+1
=...
=(3^32-1)(3^32+1)+1
=3^64-1+1
=3^64
= 3.4336838202925 * 10^30