题目
(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1=?
提问时间：2020-11-11
答案
(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3-1) * (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^2-1) * (3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^4-1) * (3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^8-1) * (3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^16-1) * (3^16+1)(3^32+1)+1
= 1/2 * (3^32-1) * (3^32+1)+1
= 1/2 * (3^64-1) + 1
= 3^64/2 - 1/2 + 1
= 3^64/2 + 1/2
= 1/2 * (3-1) * (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^2-1) * (3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^4-1) * (3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^8-1) * (3^8+1)(3^16+1)(3^32+1)+1
= 1/2 * (3^16-1) * (3^16+1)(3^32+1)+1
= 1/2 * (3^32-1) * (3^32+1)+1
= 1/2 * (3^64-1) + 1
= 3^64/2 - 1/2 + 1
= 3^64/2 + 1/2
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