SAT College

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! ​

Answers

Answer 1

Answer:

The additional words is [tex]2^n[/tex]

Explanation:

Given

[tex]B(n) = 2^n[/tex]

Required

Determine the additional words; i.e. [tex]B(n + 1) - B(n)[/tex]

From the given parameters, we have that;

B is a function of n

Such that;

[tex]B(n) = 2^n[/tex]

To calculate [tex]B(n+1)[/tex], we simply substitute n + 1 for n

[tex]B(n) = 2^n[/tex]

[tex]B(n + 1) = 2^{n + 1}[/tex]

Applying laws of indices

[tex]B(n + 1) = 2^{n} * 2^1[/tex]

[tex]B(n + 1) = 2^{n} * 2[/tex]

[tex]B(n + 1) = 2(2^{n})[/tex]

Calculating Additional Binary Code;

[tex]B(n + 1) - B(n)[/tex]

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

[tex]B(n + 1) - B(n) = 2(2^n) - 2^n[/tex]

Express [tex]2^n[/tex] as [tex]2^ n * 1[/tex]

[tex]B(n + 1) - B(n) = 2(2^n) - 2^n * 1[/tex]

Express 1 as [tex]2^0[/tex]

[tex]B(n + 1) - B(n) = 2(2^n) - 2^n * 2^0[/tex]

Factorize

[tex]B(n + 1) - B(n) = 2^n(2 - 2^0)[/tex]

[tex]B(n + 1) - B(n) = 2^n(2 - 1)[/tex]

[tex]B(n + 1) - B(n) = 2^n(1)[/tex]

[tex]B(n + 1) - B(n) = 2^n[/tex]

Hence, the additional words is [tex]2^n[/tex]

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