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

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/23 00:21:50
计算2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
反复运用公式(a-b)(a+b) = a^2-b^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)(3^8+1)(3^16+1)(3^32+1)+1
= (3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1
= (3^4-1)(3^4+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.