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

2 Answers
P dilip_k
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#

George C.
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#

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