The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n - f. The range of decimal values for X in this representation is. 

Diagram:

i represents an integral part of the and f represents the fractional part of the number.

Since, the n number is in unsigned representation, it's decimal value starts with 0. So Minimum value will be zero.

Range of unsigned representation is 0 to 2ⁱ - 1.

So, the mum value with i bits goes to 2ⁱ - 1.

Fraction of value is in the form of 2^(-i). So, when we take the value of i = 1, 2, 3 … n this range of fractional value goes like, 2^-1, 2^-2, 2^-3, …

So, it makes a GP series, with f bit maximum number possible is sum of GP series.

Consider a = ½, r = ½

Maximum value with f bits possible

= \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \ldots \)

= \(a \times \frac{{1 - {r^n}}}{{1 - r}}\)

\(= \frac{1}{2}\;\frac{{1 - {{\left( {\frac{1}{2}} \right)}^f}}}{{1 - \frac{1}{2}\;}} = 1 - {2^{ - f}}\;\)

So, maximum fractional value possible

= maximum value with i bits + maximum value with f bits

= 2ⁱ - 1 + 1 - 2^-f

= 2ⁱ - 2^-f