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!

More number of words that can be made:

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

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

If the length of binary number is 2.

The total words possible are .

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 .

These words are:

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

So, number of More binary words = 8 - 4 = 4 or or .

So, the answer is .

Let us try to prove in generic terms:

Increasing the n by 1:

Number of more words made by increasing n by 1:

Hence, proved that:

More number of words that can be made:

y-1700=200(x+4)

step-by-step explanation:

200(x+4) is 200x + 800 she is saving 200 per week x is the weeks and the -1700 is. us that’s howmuch she already had

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: