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

显示全部楼层 · 2009-3-23 22:31:57

把过程写的详细一点，谢谢啦O(∩_∩)O~

千问 · 2009-3-23 22:31:57

(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)...

千问 · 2009-3-23 22:31:57

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)...

千问 · 2009-3-23 22:31:57

回答

千问 · 2009-3-23 22:31:57

(31/3)(3^21/3^2)（3^41/3^4)(3^81/3^8）(3^161/3^16=(3-1/3)*(31/3)(3^21/3^2)（3^41/3^4)(3^81/3^8）(3^161/3^16)/(3-1/3)=(3^2-1/3^2)(3^21/3^2)（3...