Expanding the expression gives
[tex](3n+1)^2-(3n-1)^2=(9n^2+6n+1)-(9n^2-6n+1)=12n[/tex]
and clearly [tex]12\mid12n[/tex] for any natural number [tex]n[/tex].
For any natural value of n, the value of the expression is divisible by 12.
The value of n is 1, 2, 3...... n.
The numbers are positive integers and include numbers from 1 till infinity(∞).
Given
Expression; (3n+1)^2–(3n–1)^2 is divisible by 12.
Simplify the expression;
[tex]=\rm (3n+1)^2-(3n-1)^2 \\\\= 9n^2+1+6n-(9n^2+1-6n)\\\\= 9n^2+1+6n-9n^2-1+6n\\\\ = 12n[/tex]
The expression is equivalent to 12n.
When the value of n is 1.
[tex]= \dfrac{12n}{12}\\\\n=1\\\\=\dfrac{12(1)}{12}\\\\ = \dfrac{12}{12}\\\\=1[/tex]
Therefore, for any natural number value of n, the value of the expression is divisible by 12.
Hence proved.
To know more about Natural numbers click the link given below.
https://brainly.com/question/17429689
