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Find the cardinality of the following set.

[tex] n(B)=\square \quad B=\{1,3,5, \ldots, 25\} [/tex]

Enter an integer or decimal number.


Sagot :

To find the cardinality of the set \( B = \{1, 3, 5, \ldots, 25\} \), we need to determine the number of elements in the set. Let’s examine the sequence within the set \( B \):

1. The set \( B \) consists of all odd numbers starting from 1 and going up to 25.

2. The sequence of odd numbers can be described as \( 1, 3, 5, 7, \ldots, 25 \).

3. We can recognize that this set forms an arithmetic sequence with the first term \( a_1 = 1 \) and a common difference \( d = 2 \).

4. The general formula for the \( n \)-th term of an arithmetic sequence is given by:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]

5. We need to find the value of \( n \) for which \( a_n = 25 \):
[tex]\[ 25 = 1 + (n-1) \cdot 2 \][/tex]
Simplifying this equation:
[tex]\[ 25 = 1 + 2n - 2 \][/tex]
[tex]\[ 25 = 2n - 1 \][/tex]
[tex]\[ 26 = 2n \][/tex]
[tex]\[ n = 13 \][/tex]

Thus, there are 13 terms in the sequence \( B \), so the cardinality of the set \( B \) is:

[tex]\[ n(B) = 13 \][/tex]

Therefore, the cardinality of the set [tex]\( B \)[/tex] is [tex]\( \boxed{13} \)[/tex].

Answer:

13

Step-by-step explanation:

The cardinality of B = {1,3,5,7,9,11,13,15,17,19,21,23,25} = n(B) = the number of elements that B contains = 13