您的位置: 首页  >  作业答案  >  数学  >  计算(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)-1

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

分类: 作业答案 • 2021-12-19 12:05:01
问题描述：

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

答

原式=[(3-1)*(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/(3-1)]-1
=[(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/(3-1)]-1
=[(3^32-1)/2]-1
=3/2(3^31-1)

答

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

相关推荐