Question

Following is a subset of:
\[\{n\left( \left( n+1 \right)\left( 2n+1 \right) \right):n\] belongs to Z \[\}\]
(1) \[\{6K:K\] belongs to Z \[\}\]
(2) \[\{12K:K\] belongs to Z \[\}\]
(3) \[\{18K:K\] belongs to Z \[\}\]
(4) \[\{24K:K\] belongs to Z \[\}\]

Answer 1

Hint: We are given a set and we are asked to find the superset of which the given set is the subset from the given options. We will first find the different values that the given set can for values of \[n\] belonging to the set of integers, Z. Then, we will compare the values to the sets given in the options for the set that can produce those values. Hence, we will have a super set of the given set.

Complete step by step answer:
According to the given question, we are given a set and we are asked in the question to find the superset of the given set.
The given set that we have is,
\[\{n\left( \left( n+1 \right)\left( 2n+1 \right) \right):n\] belongs to Z \[\}\]
Since, \[n\] belongs to the set of integers, that is, \[n\] can have all integer values both negative and positive numbers.
We will calculate the values that the given set can have by substituting the values of \[n\].
Let us suppose that the given set be equal to,
\[A=\{n\left( \left( n+1 \right)\left( 2n+1 \right) \right):n\in Z\}\]
Now, substituting values of \[n\] we have the set as,
\[n=\{...-3,-2,-1,0,1,2...\}\]
We get the values in the set as,
\[A=\{...-30,-6,0,0,6,30...\}\]
We can see that the given set can have values like 6 and its multiples, so the most suitable super set for the given set is \[\{6K:K\] belongs to Z \[\}\]
Therefore, \[\{n\left( \left( n+1 \right)\left( 2n+1 \right) \right):n\] belongs to Z \[\}\] is the subset of option (1) \[\{6K:K\] belongs to Z \[\}\].

So, the correct answer is “Option 1”.

Note: The other options we are given may look suitable to be the superset of the given set, but the values that we obtained for the given set can only be obtained from the set \[\{6K:K\] belongs to Z \[\}\] as the value 6 cannot be obtained by other sets where \[n\] belongs to \[Z\].