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:

The additional words is $2^n$

Explanation:

Given

$B(n) = 2^n$

Required

Determine the additional words; i.e. $B(n + 1) - B(n)$

From the given parameters, we have that;

B is a function of n

Such that;

$B(n) = 2^n$

To calculate $B(n+1)$, we simply substitute n + 1 for n

$B(n) = 2^n$

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

Applying laws of indices

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

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

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

Calculating Additional Binary Code;

$B(n + 1) - B(n)$

Substitute values for B(n + 1) and B(n)

$B(n + 1) - B(n) = 2(2^n) - 2^n$

Express $2^n$ as $2^ n * 1$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 1$

Express 1 as $2^0$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 2^0$

Factorize

$B(n + 1) - B(n) = 2^n(2 - 2^0)$

$B(n + 1) - B(n) = 2^n(2 - 1)$

$B(n + 1) - B(n) = 2^n(1)$

$B(n + 1) - B(n) = 2^n$

Hence, the additional words is $2^n$