((2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1)/(2^33)=?

2 Answers
Jul 30, 2018

((2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1)/(2^33)

=1/2^2((1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16))+1/2^33

=1/2^2(1+1/2^2+1/2^3+1/2^4+cdot+1/2^31)+1/2^33

=1/2^2[(1*(1-1/2^32))/(1-1/2)]+1/2^33

=1/2(1-1/2^32)+1/2^33

=1/2-1/2^33+1/2^33

=1/2

Jul 30, 2018

1/2

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