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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. If the length is increased from
n to
n+1, how many more binary code words will there be?
Choose 1 answer:
(A) 2
(B)
2^(n)
(C)
2^(n+1)
(D)
4^(n)

$B(n)=2^{n}$ $\newline$ 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. If the length is increased from $n$ to $n+1$ , how many more binary code words will there be? $\newline$ Choose $1$ answer: $\newline$ (A) $2$ $\newline$ (B) $2^{n}$ $\newline$ (C) $2^{n+1}$ $\newline$ (D) $4^{n}$

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Algebra 1

Compare linear and exponential growth

Full solution

B(n)=2^{n}

\newline

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. If the length is increased from

n

n+1

, how many more binary code words will there be?

\newline

Choose

1

answer:

\newline

(A)

2

\newline

(B)

2^{n}

\newline

(C)

2^{n+1}

\newline

(D)

4^{n}

Problem Understanding: Understand the problem. $\newline$ We are given a function $B(n) = 2^n$ , which represents the number of binary code words of length $n$ . We need to find out how many additional binary code words are possible when the length increases from $n$ to $n+1$ .
Calculate Binary Code Words of Length $n$ : Calculate the number of binary code words of length $n$ . Using the given function, the number of binary code words of length $n$ is $B(n) = 2^n$ .
Calculate Binary Code Words of Length $n+1$ : Calculate the number of binary code words of length $n+1$ . Using the same function, the number of binary code words of length $n+1$ is $B(n+1) = 2^{(n+1)}$ .
Determine Increase in Binary Code Words: Determine the increase in the number of binary code words. $\newline$ To find out how many more binary code words there are when the length increases from $n$ to $n+1$ , we subtract the number of binary code words of length $n$ from the number of binary code words of length $n+1$ . $\newline$ So, the increase is $B(n+1) - B(n) = 2^{(n+1)} - 2^n$ .
Simplify Expression for Increase: Simplify the expression for the increase. $\newline$ We can factor out $2^n$ from the expression to simplify it: $\newline$ $B(n+1) - B(n) = 2^n \times (2 - 1) = 2^n \times 1 = 2^n$ .
Choose Correct Answer: Choose the correct answer. $\newline$ The increase in the number of binary code words when the length increases from $n$ to $n+1$ is $2^n$ . This corresponds to answer choice (B) $2^n$ .