((2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1)/(2^33)=?
2 Answers
Jul 30, 2018
Jul 30, 2018
Explanation:
Here's one method...
((2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1)/2^33
=(2^31+2^30+2^29+...+2^3+2^2+2^1+2^0+2^0)/2^33
=(2^31+2^30+2^29+...+2^3+2^2+2^1+2^1)/2^33
=(2^31+2^30+2^29+...+2^3+2^2+2^2)/2^33
vdots
=(2^31+2^31)/2^33
=2^32/2^33
=1/2
Or if you prefer, let's use binary...
((2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1)/2^33
=(2^31+2^30+2^29+...+2^3+2^2+2^1+2^0+2^0)/2^33
=(overbrace(111...1_2)^"32 digits"+1)/(underbrace(100...0_2)_"34 digits")
=(overbrace(100...0_2)^"33 digits")/(underbrace(100...0_2)_"34 digits")
=1/2