Answer:
More number of words that can be made: [tex]\bold{2^n}[/tex]
Please refer to below proof.
Step-by-step explanation:
Given that:
The number of binary code words that can be made:
[tex]B(n) =2^n[/tex]
where n is the length of binary numbers.
Binary numbers means 2 possibilities either 0 or 1.
Here, suppose if we have 5 as the length of binary number.
And there are 2 possibilities for each digit.
So, total number of possibilities will be [tex]2\times 2\times 2\times 2\times 2 = 2^5[/tex]
If the length of binary number is 2.
The total words possible are [tex]2^2[/tex].
These numbers are:
{00, 01, 10, 11}
If the length of binary number is 3. (increasing the 'n' by 1)
The total words possible are [tex]2^3[/tex].
These words are:
{000, 001, 010, 100, 011, 101, 110, 111}
So, number of More binary words = 8 - 4 = 4 or [tex]2^2[/tex] or [tex]2^n[/tex].
So, the answer is [tex]2^n[/tex].
Let us try to prove in generic terms:
[tex]B(n) = 2^n[/tex]
Increasing the n by 1:
[tex]B(n+1) = 2^{n+1}[/tex]
Number of more words made by increasing n by 1:
[tex]B(n+1) -B(n)= 2^{n+1} -2^n\\\Rightarrow 2\times 2^{n} -2^n\\\Rightarrow 2^n(2-1)\\\Rightarrow \bold{2^n}[/tex]
Hence, proved that:
More number of words that can be made: [tex]\bold{2^n}[/tex]
