Answer:
Explanation:
From the given information:
[tex]a_{n-1} , a_{n-2}...a_o[/tex] in binary is:
[tex]a_{n-1}\times 2^{n-1} + a_{n-2}}\times 2^{n-2}+ ...+a_o[/tex]
So, the largest number posses all [tex]a_{n-1} , a_{n-2}...a_o[/tex] nonzero, however, the smallest number has [tex]a_{n-2} , a_{n-3}...a_o[/tex] all zero.
∴
The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times
i.e.
[tex](11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}[/tex]
[tex]= \dfrac{1(2^n-1)}{2-1}[/tex]
[tex]\mathbf{=2^n -1}[/tex]
[tex](1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}[/tex]
[tex]\mathbf {= 2 ^{n-1}}[/tex]
Hence, the smallest value is [tex]\mathbf{2^{n-1}}[/tex] and the largest value is [tex]\mathbf{2^{n}-1}[/tex]
