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3 years ago

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Vaselesa [24]3 years ago

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Yes i is a time to be like the only one who me haha was the way i is a good day

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Soloha48 [4]

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(b) 31.85

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4 years ago

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 len

Citrus2011 [14]

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 $2^n$

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3 years ago