0 votes 0 votes if we have $n$ consecutive $1s$ in binary then its magnitude is $2^n -1$ similarly ,do we have any short cut like ,if we have $n$ consecutive $1s$ after a decimal point then the magnitude will be $1-2^n$ ? Digital Logic easy digital-logic number-representation + – HHH777 asked Mar 25, 2018 edited Mar 29, 2018 by Sukanya Das HHH777 535 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes 1111111.........( n times) = 2^n-1 (magnitude) 0.11111..........( n times) = 1-2^-n (magnitude) abhishekmehta4u answered Mar 25, 2018 abhishekmehta4u comment Share Follow See 1 comment See all 1 1 comment reply pilluverma123 commented Mar 25, 2018 reply Follow Share We can just use G.P. series sum - $2^{-1} + 2^{-2} + ....... + 2^{-n} = \frac{\left ( \frac{1}{2}\left ( 1 - 2^{-n} \right ) \right )}{\frac{1}{2}} = 1 - 2^{-n}$ 3 votes 3 votes Please log in or register to add a comment.