Question

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of length 4. The number of binary code words, B(n), of length n, is shown above. If the length is increased from n to n+1, how many more binary code words will there be? The answer is 2^n, but I don't get how they got that answer. I would think 2^n+1 minus 2^n would be 2. Please help me! Thank you!

Answer 1

Answer:

More number of words that can be made: $\bold{2^n}$

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

$B(n) =2^n$

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 $2\times 2\times 2\times 2\times 2 = 2^5$

If the length of binary number is 2.

The total words possible are $2^2$.

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 $2^3$.

These words are:

{000, 001, 010, 100, 011, 101, 110, 111}

So, number of More binary words = 8 - 4 = 4 or $2^2$ or $2^n$.

So, the answer is $2^n$.

Let us try to prove in generic terms:

$B(n) = 2^n$

Increasing the n by 1:

$B(n+1) = 2^{n+1}$

Number of more words made by increasing n by 1:

$B(n+1) -B(n)= 2^{n+1} -2^n\\\Rightarrow 2\times 2^{n} -2^n\\\Rightarrow 2^n(2-1)\\\Rightarrow \bold{2^n}$

Hence, proved that:

More number of words that can be made: $\bold{2^n}$