I am having trouble with this question.
$$B(n)=2^n$$
A binary codeword of length $n$ is a string of 000's and 111'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?
I know that I will have to find the difference between the original equation of $2^n$ and $2^{n+1}$ but I don't understand how to do that.
Thank you for your help!
If you change $n$ to $n+1$, the number of codes doubles.
Note that $$2^{n+1} = 2(2^n)$$
for example if $n=5$, $2^n= 2^5 =32.$
If we change $5$ to $6$ we get $ 2^6=64$ which is twice $32.$