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

The additional words is
`2^n`

Step-by-step 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`