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RosaJackson8088
05.08.2020 •
Mathematics
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!
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Ответ:
More number of words that can be made:![\bold{2^n}](/tpl/images/0717/5560/b868e.png)
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![2\times 2\times 2\times 2\times 2 = 2^5](/tpl/images/0717/5560/9bf84.png)
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:![\bold{2^n}](/tpl/images/0717/5560/b868e.png)
Ответ: