计算(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)-3^64/2要详细过程急!
问题描述：
计算(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)-3^64/2
要详细过程
急!
答
把每个跨号因子变下,如(a^n)+1=((a^2n)-1)/((a^n)-1)
这样每相连2个因子变成可以前分子后分母相消去的项
这样前面得到结果((3^64)-1)/2
后面相减,结果-1/2
答
(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)-3^64/2
=[(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)]/2-3^64/2
=[(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)]/2-3^64/2
=[(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)]/2-3^64/2
=[(3^8-1)(3^8+1)(3^16+1)(3^32+1)]/2-3^64/2
=[(3^16-1)(3^16+1)(3^32+1)]/2-3^64/2
=[(3^32-1)(3^32+1)]/2-3^64/2
=(3^64-1)/2-3^64/2
=3^64/2-1/2-3^64/2
=-1/2
答
原式=（3-1）(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)-3^64/2
=(3^2-1))(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)/2-3^64/2
=（3^32-1)(3^32+1)/2-3^64/2
=3^64/2-1/2-3^64/2
=-1/2